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✓ 22 results · all verified · 9 also independently AI-judged

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Analytic Semigroups and Linear Evolution Equations

1 · Prerequisites

2 · Summary

This page develops analytic semigroups from the complex-analytic calculus that produces them through the PDE realisations that motivate them. The Banach-valued Cauchy and Taylor machinery supplies primitives, power-series expansions and Cauchy estimates for holomorphic curves; these are used to define sectorial operators, to construct the Dunford contour family and to prove its semigroup law, and to identify the generator of the contour semigroup with the original operator. The smoothing estimates and their operator-norm consequences show that positive-time orbits land in every graph domain $D (A^{m})$ with the standard $t^{- m}$ singularity, while the sectorial-resolvent characterisation turns resolvent bounds into bounded analytic generation and back.

Two complementary routes then generate semigroups from quadratic data. Self-adjoint nonpositive operators give contraction analytic semigroups of angle $π / 2$ , with the spectral-gap corollary yielding exponential decay, and closed sectorial forms with the stated norm comparison give analytic semigroups on every sector strictly narrower than $π / 2 - θ$ , where $θ$ is the form half-angle, with the Dirichlet Laplacian as the principal PDE specialisation. The final block treats the classical evolution problem: the Duhamel cancellation of the generator singularity, the domain compatibility $x \in D (A)$ and endpoint identity $u^{'} (0) = A x + f (0)$ , and classical regularity for Hölder continuous forcing, together with an abstract smoothing corollary and the remark separating abstract graph-domain smoothing from its spatial reading.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Resolvent identity and holomorphy for a closed operator

Statement

Let $X$ be a Banach space over $K \in {R, C}$ (Banach space, Real and complex scalar conventions for normed spaces) and let $A : D (A) \subseteq X \to X$ be a closed linear operator with resolvent $R (λ, A) = (λ I - A)^{- 1} \in B (X)$ for $λ \in ρ (A)$ (Resolvent and spectrum of a closed operator on a Banach space). Then:

$ρ (A)$ is open: if $λ_{0} \in ρ (A)$ and $∣ λ - λ_{0} ∣ ∥ R (λ_{0}, A) ∥ < 1$ , then $λ \in ρ (A)$ and $R (λ, A) = \sum_{n \geq 0} (λ_{0} - λ)^{n} R (λ_{0}, A)^{n + 1},$ the series converging in the operator norm;
the resolvent identity $R (λ, A) - R (μ, A) = (μ - λ) R (λ, A) R (μ, A) = (μ - λ) R (μ, A) R (λ, A)$ holds for all $λ, μ \in ρ (A)$ ;
$λ \mapsto R (λ, A)$ is differentiable on $ρ (A)$ in the operator norm with derivative $R^{'} (λ, A) = - R (λ, A)^{2}$ ; when $K = C$ this is norm-holomorphy (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions);
for every $z \in C$ the map $λ \mapsto e^{λ z} R (λ, A)$ is differentiable on $ρ (A)$ in the operator norm with derivative $z e^{λ z} R (λ, A) - e^{λ z} R (λ, A)^{2}$ , and norm-holomorphic when $K = C$ ; for $K = R$ the complex scalar $e^{λ z}$ acts through the canonical complexification of $X$ (Canonical Banach complexification of a real Banach space).

No choice principle is used.

Facts & Assumptions

Given: A Banach space $X$ over $K \in {R, C}$ , a closed linear operator $A : D (A) \subseteq X \to X$ , its resolvent set $ρ (A)$ , the resolvents $R (λ, A) = (λ I - A)^{- 1}$ for $λ \in ρ (A)$ , and the operator $T : = (λ_{0} - λ) R (λ_{0}, A)$ attached to a fixed $λ_{0} \in ρ (A)$ and $λ \in K$ .

[L1]

For $σ \in ρ (A)$ one has $R (σ, A) \in B (X)$ , $R (σ, A) X = D (A)$ , $R (σ, A) (σ I - A) y = y$ for $y \in D (A)$ , $(σ I - A) R (σ, A) x = x$ for $x \in X$ , and $A R (σ, A) = σ R (σ, A) - I \in B (X)$ (Resolvent and spectrum of a closed operator on a Banach space).

[L2]

If $R \in B (X)$ satisfies $∥ R ∥ < 1$ , then $I - R$ is invertible with inverse the operator-norm limit $\sum_{n \geq 0} R^{n}$ , and $∥ \sum_{n \geq 0} R^{n} ∥ \leq (1 - ∥ R ∥)^{- 1}$ (Neumann series and small perturbations of bounded inverses); moreover $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ for the operator norm (Composition satisfies |ST|\le|S|,|T|).

[L3]

The complex exponential is entire with $\exp^{'} (z) = \exp (z)$ (The complex exponential is entire and its complex derivative is itself), and its defining series gives $\exp (0) = 1$ (The complex exponential by its power series); hence for every fixed $z \in C$ the difference quotient $h^{- 1} (e^{h z} - 1) \to z$ as $h \to 0$ in $C$ , because $e^{h z} = O (1)$ -bounded near $0$ and $h^{- 1} (e^{h z} - 1) = z \cdot (h z)^{- 1} (\exp (h z) - \exp (0)) \to z \exp^{'} (0)$ .

Proof

technique · direct

1.1L1givenalgebra

Factorization. Fix $λ_{0} \in ρ (A)$ and set $T : = (λ_{0} - λ) R (λ_{0}, A) \in B (X)$ . For every $y \in D (A)$ , writing $z : = (λ_{0} I - A) y$ gives $y = R (λ_{0}, A) z$ by [L1] and $(λ I - A) y = (λ_{0} I - A) y + (λ - λ_{0}) y = z + (λ - λ_{0}) R (λ_{0}, A) z = (I - T) (λ_{0} I - A) y,$ so $λ I - A = (I - T) (λ_{0} I - A)$ as maps $D (A) \to X$ .

1.2L1givenalgebra

Resolvent identity. For $λ, μ \in ρ (A)$ the identity $(μ - λ) R (μ, A) = (μ I - A) R (μ, A) - (λ I - A) R (μ, A) = I - (λ I - A) R (μ, A)$ holds on $X$ , because both resolvents are everywhere defined and $(μ I - A) R (μ, A) = I$ by [L1]. Hence, using $R (μ, A) X \subseteq D (A)$ and $R (λ, A) (λ I - A) z = z$ for $z \in D (A)$ , $(μ - λ) R (λ, A) R (μ, A) = R (λ, A) (I - (λ I - A) R (μ, A)) = R (λ, A) - R (μ, A) .$ Exchanging $λ$ and $μ$ gives $(λ - μ) R (μ, A) R (λ, A) = R (μ, A) - R (λ, A)$ ; combining the two displays yields the second form $(μ - λ) R (λ, A) R (μ, A) = (μ - λ) R (μ, A) R (λ, A)$ for $μ \neq λ$ , while for $μ = λ$ both sides vanish.

2.1step 1.1L1L2algebra

Openness and the expansion. If $∥ T ∥ = ∣ λ - λ_{0} ∣ ∥ R (λ_{0}, A) ∥ < 1$ , then [L2] makes $I - T$ invertible with inverse $S : = \sum_{n \geq 0} T^{n}$ . For $x \in X$ put $y : = R (λ_{0}, A) S x \in D (A)$ ; by [step 1.1] and [L1], $(λ I - A) y = (I - T) (λ_{0} I - A) R (λ_{0}, A) S x = (I - T) S x = x,$ so $λ I - A$ is surjective; it is injective because $(λ I - A) y = 0$ and [step 1.1] give $(I - T) (λ_{0} I - A) y = 0$ , hence $(λ_{0} I - A) y = 0$ and $y = R (λ_{0}, A) 0 = 0$ . Thus $λ \in ρ (A)$ and $R (λ, A) = R (λ_{0}, A) (I - T)^{- 1} = R (λ_{0}, A) \sum_{n \geq 0} ((λ_{0} - λ) R (λ_{0}, A))^{n} = \sum_{n \geq 0} (λ_{0} - λ)^{n} R (λ_{0}, A)^{n + 1},$ the series converging in operator norm because $∥ (λ_{0} - λ)^{n} R (λ_{0}, A)^{n + 1} ∥ \leq ∥ R (λ_{0}, A) ∥ ∥ T ∥^{n}$ and $∥ T ∥ < 1$ .

3.1step 2.1L2algebra

Differentiability of the resolvent. Let $λ_{0} \in ρ (A)$ and let $h \in K$ with $∣ h ∣ ∥ R ∥ < 1$ , where $R : = R (λ_{0}, A)$ . By [step 2.1] applied to the pair $λ_{0} + h, λ_{0}$ , $R (λ_{0} + h, A) - R (λ_{0}, A) = \sum_{n \geq 1} (- h)^{n} R^{n + 1} = - h R^{2} + h^{2} \sum_{n \geq 2} (- h)^{n - 2} R^{n + 1},$ and the norm of the second summand is at most $∣ h ∣^{2} ∥ R ∥^{3} (1 - ∣ h ∣ ∥ R ∥)^{- 1}$ , so $∥ h^{- 1} (R (λ_{0} + h, A) - R (λ_{0}, A)) + R^{2} ∥ \leq ∣ h ∣ ∥ R ∥^{3} (1 - ∣ h ∣ ∥ R ∥)^{- 1} \to 0$ . Hence $λ \mapsto R (λ, A)$ is differentiable at $λ_{0}$ with derivative $- R (λ_{0}, A)^{2}$ ; when $K = C$ this is complex differentiability in operator norm, that is, norm-holomorphy.

3.2step 2.1L2algebra

Local boundedness and continuity. With $R : = R (λ_{0}, A)$ and $∣ h ∣ ∥ R ∥ < 1$ , the expansion of [step 2.1] gives $∥ R (λ_{0} + h, A) - R (λ_{0}, A) ∥ \leq \sum_{n \geq 1} ∣ h ∣^{n} ∥ R ∥^{n + 1} = \frac{∣ h ∣ ∥ R ∥^{2}}{1 - ∣ h ∣ ∥ R ∥} ⟶ 0$ and $∥ R (λ_{0} + h, A) ∥ \leq ∥ R ∥ (1 - ∣ h ∣ ∥ R ∥)^{- 1}$ ; thus $λ \mapsto R (λ, A)$ is continuous at every point of $ρ (A)$ and locally bounded in operator norm.

4.1step 3.1step 3.2L3algebra

The exponential factor. Fix $λ_{0} \in ρ (A)$ and $z \in C$ , write $R = R (λ_{0}, A)$ , and let $h \neq 0$ with $∣ h ∣ ∥ R ∥ < 1$ . Then $\frac{e^{(λ_{0} + h) z} R (λ_{0} + h, A) - e^{λ_{0} z} R (λ_{0}, A)}{h} = e^{λ_{0} z} \frac{e^{h z} - 1}{h} R (λ_{0} + h, A) + e^{λ_{0} z} \frac{R (λ_{0} + h, A) - R (λ_{0}, A)}{h} .$ As $h \to 0$ the first factor $\frac{e^{h z} - 1}{h} \to z$ by [L3], the second factor $R (λ_{0} + h, A) \to R (λ_{0}, A)$ in operator norm by [step 3.2], and the last difference quotient tends to $- R^{2}$ by [step 3.1]; multiplying by the bounded scalars $e^{λ_{0} z}$ gives convergence in operator norm to $z e^{λ_{0} z} R (λ_{0}, A) - e^{λ_{0} z} R (λ_{0}, A)^{2}$ .

5.1step 1.2step 2.1step 3.1step 4.1given∎

Collecting [step 2.1] (openness and the displayed expansion, claim 1), [step 1.2] (the resolvent identity, claim 2), [step 3.1] (norm differentiability with derivative $- R^{2}$ , and norm-holomorphy over $C$ , claim 3) and [step 4.1] (the exponential factor, claim 4) proves the four claims; every step used only the resolvent identities, the Neumann expansion and the scalar exponential, so no choice principle was used.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited integral and semigroup suppliers.

Let $Y$ be a complex Banach space (Banach space), let $U \subseteq C$ be open and star-shaped with base point $a \in U$ (so $[a, z] \subseteq U$ for every $z \in U$ ; A complex domain is a nonempty connected open subset of $C$ ), and let $F : U \to Y$ be continuous and complex-differentiable on $U$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions). For a piecewise $C^{1}$ contour $γ : [α, β] \to U$ put $\int_{γ} F d w : = \int_{α}^{β} F (γ (t)) γ^{'} (t) d t,$ a Bochner integral (Bochner-integrable function), the Banach-valued analogue of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral. Then:

the segment integral $G (z) : = \int_{0}^{1} F (a + t (z - a)) (z - a) d t$ is a well-defined element of $Y$ for every $z \in U$ , and $G : U \to Y$ is complex-differentiable with $G^{'} (z) = F (z)$ on $U$ ;
for every closed piecewise $C^{1}$ contour $γ : [α, β] \to U$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation) one has $\int_{γ} F d w = 0$ , and for two such contours in $U$ with common initial and terminal point the integrals agree.

No choice principle beyond Countable Choice is used.

Facts & Assumptions

Given: An open star-shaped $U \subseteq C$ with base point $a$ , a continuous complex-differentiable $F : U \to Y$ into a complex Banach space $Y$ , and the segment integral $G (z) = \int_{0}^{1} F (a + t (z - a)) (z - a) d t$ .

[L1]

A continuous $f : [u, v] \to Y$ is Bochner integrable and its primitive is differentiable with derivative $f$ ; for a curve $φ$ continuous on $[u, v]$ , differentiable in the interior with derivative extending continuously, $\int_{u}^{v} φ^{'} = φ (v) - φ (u)$ (Fundamental theorem of calculus for Banach-valued continuous curves).

[L2]

The Bochner integral is linear in the integrand and $∥ \int_{E} f ∥ \leq \int_{E} ∥ f ∥$ (Linearity of the Bochner integral, Bochner integral norm inequality).

[L3]

A contour is a rectifiable path; it is closed when its endpoints agree, its reversal is $γ^{-} (t) = γ (a + b - t)$ , and concatenation $α * β$ is defined when $α (1) = β (0)$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

Proof

technique · direct

1.1L1L2L3givenconstruct

Triangle subdivision. For a closed nondegenerate triangle $Δ \subset U$ , put $I (Δ) = \int_{\partial Δ} F d w$ . Subdivide into four similar triangles, with matching boundary orientations; internal edges cancel by [L2, L3], so some child has integral norm at least $∥ I (Δ) ∥ / 4$ . Order the four children once and take the first satisfying this inequality at each subdivision. The resulting nested triangles $Δ_{n}$ have diameter $2^{- n} d$ and perimeter $2^{- n} p$ , where $d, p$ are those of $Δ$ , and $∥ I (Δ_{n}) ∥ \geq 4^{- n} ∥ I (Δ) ∥$ . Their intersection is a point $w_{0}$ : a specified vertex of each triangle is a Cauchy sequence in $C$ , its limit lies in every closed triangle, and the diameters tend to zero.

2.1step 1.1L1L2L3givenalgebra

Goursat's estimate. Differentiability at $w_{0}$ gives $F (w) = F (w_{0}) + F^{'} (w_{0}) (w - w_{0}) + (w - w_{0}) r (w)$ with $r (w) \to 0$ as $w \to w_{0}$ and $r (w_{0}) = 0$ . The affine part has polynomial primitive $F (w_{0}) w + F^{'} (w_{0}) (w - w_{0})^{2} / 2$ , so its boundary integral vanishes by [L1]. On $Δ_{n}$ , $∣ w - w_{0} ∣ \leq 2^{- n} d$ and $\sup_{Δ_{n}} ∥ r ∥ \to 0$ ; hence [L2] gives $∥ I (Δ_{n}) ∥ \leq 4^{- n} p d \sup_{Δ_{n}} ∥ r ∥$ . Comparing with step 1.1 proves $I (Δ) = 0$ . For a degenerate triangle the oriented segment integrals cancel directly.

3.1step 2.1L1L2L3givenalgebra

The segment primitive. The segment integrand defining $G (z)$ is continuous, so [L1] makes it integrable. Fix $z \in U$ and take $h$ sufficiently small that $[z, z + h] \subset U$ . Every point of $conv {a, z, z + h}$ is on a segment from $a$ to a point of $[z, z + h]$ , so the triangle lies in $U$ . Its boundary integral is zero by step 2.1; additivity and reversal therefore give $G (z + h) - G (z) = \int_{[z, z + h]} F d w = h \int_{0}^{1} F (z + t h) d t$ . The norm of the difference between this quotient and $F (z)$ is at most $\sup_{0 \leq t \leq 1} ∥ F (z + t h) - F (z) ∥$ , which tends to zero by continuity. Thus $G^{'} = F$ .

4.1step 3.1L1L2L3givenalgebra∎

Closed contours. On each $C^{1}$ piece of a contour $γ$ , the difference-quotient chain rule gives $(G \circ γ)^{'} = F (γ) γ^{'}$ ; this derivative is continuous on the closed piece because $F$ and $γ^{'}$ are continuous. Applying [L1] piecewise and telescoping gives $\int_{γ} F d w = G (γ (β)) - G (γ (α))$ . It is zero for a closed contour; concatenating a contour with the reversal of another having the same endpoints gives path independence. The subdivision choices were specified by a finite ordering, so no choice principle beyond Countable Choice was used.

Remarks

The triangle-subdivision argument uses the differentiability remainder only on triangles shrinking to its base point. It does not estimate that remainder on a fixed segment from the star center.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Cauchy integral formula and Cauchy estimates for Banach-valued holomorphic functions

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited integral and semigroup suppliers.

Let $Y$ be a complex Banach space (Banach space), let $U \subseteq C$ be open (A complex domain is a nonempty connected open subset of $C$ ), and let $F : U \to Y$ be continuous and complex-differentiable on $U$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions). Suppose $R > 0$ and that the closed disc $\overline{D (z_{0}, R)} = {w : ∣ w - z_{0} ∣ \leq R}$ is contained in $U$ . For $0 < r < R$ write $C_{r}$ for the positively oriented circle $∣ w - z_{0} ∣ = r$ and $\oint_{C_{r}} Φ (w) d w : = \int_{0}^{2 π} Φ (z_{0} + r e^{i t}) r i e^{i t} d t,$ a Bochner integral (Bochner-integrable function, The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral). Then:

for every $r$ with $0 < r < R$ and every $z$ with $∣ z - z_{0} ∣ < r$ the Cauchy integral formula holds: $F (z) = \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = r} \frac{F (w)}{w - z} d w;$
$F$ has norm-convergent power-series expansions about $z_{0}$ on $D (z_{0}, R)$ , with $a_{n} = \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = r} \frac{F (w)}{(w - z_{0})^{n + 1}} d w$ for every $r$ with $0 < r < R$ ; these coefficient integrals are independent of $r$ ;
$F$ is norm- $C^{\infty}$ , and with $M (r) : = \sup_{∣ w - z_{0} ∣ = r} ∥ F (w) ∥$ one has the Cauchy estimates $∥ F^{(n)} (z_{0}) ∥ \leq n! \frac{M (r)}{r^{n}} (n \geq 0, 0 < r < R) .$

No choice principle beyond Countable Choice is used.

Facts & Assumptions

Given: A complex Banach space $Y$ , an open $U \subseteq C$ , a continuous complex-differentiable $F : U \to Y$ , a closed disc $\overline{D (z_{0}, R)} \subseteq U$ , numbers $0 < r < R$ , a point $z$ with $∣ z - z_{0} ∣ < r$ , the positively oriented circle $C_{r}$ with its Bochner parametrization $γ (s) = z_{0} + r e^{i s}$ , and $M (r) = \sup_{C_{r}} ∥ F ∥$ .

[L1]

The disc $D (z_{0}, R)$ is convex, hence star-shaped with base point $z_{0}$ ; by Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains, $F$ has a primitive $G$ on $D (z_{0}, R)$ with $G^{'} = F$ , every closed piecewise $C^{1}$ contour in $D (z_{0}, R)$ has $\int_{γ} F d w = 0$ , and the same supplier applies to any holomorphic map on a smaller open disc. In particular, $\oint_{C_{ρ}} F d w = 0$ for every $0 < ρ < R$ .

[L2]

The Bochner integral is linear in the integrand and $∥ \int_{E} f ∥ \leq \int_{E} ∥ f ∥$ (Linearity of the Bochner integral, Bochner integral norm inequality); closed contours and their reversals and concatenations are those of Rectifiable complex contours, reversal, concatenation, closedness, and orientation.

[L3]

If two $Y$ -valued power series $\sum a_{n} ζ^{n}$ and $\sum b_{n} ζ^{n}$ converge on a disc and their sums agree at every real point of that disc, then $a_{n} = b_{n}$ for all $n$ (Banach-valued power series are determined by their real values, Series and absolute convergence in a normed space).

Proof

technique · direct

1.1L1L2givenconstruct

The filled quotient. Put $h (w) = (F (w) - F (z)) / (w - z)$ for $w \neq z$ and $h (z) = F^{'} (z)$ . Differentiability of $F$ at $z$ makes $h$ continuous on $D (z_{0}, R)$ , and the quotient rule makes it holomorphic away from $z$ . The triangle-subdivision argument of Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains proves that a holomorphic map has zero integral around every closed triangle in its domain. This also holds for $h$ on triangles containing $z$ : split such a triangle into at most three triangles with vertex $z$ ; in each remove a similar corner triangle of diameter $η$ . The remaining quadrilateral can be split into triangles avoiding $z$ , whose integrals vanish. Its boundary differs from the original by edges of total length $O (η)$ , and $h$ is bounded near $z$ , so the norm of this difference tends to zero by [L2]. Thus every triangle integral of $h$ in the disc vanishes. Degenerate triangles cancel by reversal.

1.2L2givenalgebra

Scalar circle integrals. Parametrizing $γ (s) = z_{0} + r e^{i s}$ and setting $ρ : = z - z_{0}$ with $∣ ρ ∣ < r$ , the geometric series $\frac{1}{w - z} = \frac{1}{w - z_{0}} \sum_{n \geq 0} (\frac{z - z_{0}}{w - z_{0}})^{n} = \sum_{n \geq 0} \frac{(z - z_{0})^{n}}{(w - z_{0})^{n + 1}}$ converges uniformly on $C_{r}$ ; integrating termwise and using $\frac{1}{2 π i} \oint_{C_{r}} (w - z_{0})^{m} d w = 1$ for $m = - 1$ and $= 0$ for integers $m \neq - 1$ (a direct computation from $\int_{0}^{2 π} e^{i k t} d t = 2 π$ for $k = 0$ and $0$ otherwise) gives $\frac{1}{2 π i} \oint_{C_{r}} \frac{d w}{w - z} = 1$ and $\oint_{C_{r}} d w = 0$ .

2.1step 1.1L1L2givenalgebra

A primitive for the filled quotient. Define $H (w) = \int_{[z_{0}, w]} h (ζ) d ζ$ on the disc. The zero triangle integrals in step 1.1 give $H (w + k) - H (w) = k \int_{0}^{1} h (w + t k) d t$ for small $k$ . Continuity of $h$ gives $H^{'} = h$ , exactly as in the segment-primitive argument of Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains. Applying its piecewise chain-rule and fundamental-theorem argument to $H$ along $C_{r}$ yields $\oint_{C_{r}} h (w) d w = 0$ . This uses continuity at the exceptional point, without assuming that $h$ is differentiable there.

3.1step 1.2step 2.1L2givenalgebra

Cauchy's integral formula. Put $J : = \oint_{C_{r}} h (w) d w = 0$ . Writing $\frac{F (w)}{w - z} = \frac{F (z)}{w - z} + \frac{F (w) - F (z)}{w - z}$ and using [step 2.1] and [step 1.2], $\frac{1}{2 π i} \oint_{C_{r}} \frac{F (w)}{w - z} d w = \frac{F (z)}{2 π i} \oint_{C_{r}} \frac{d w}{w - z} + \frac{J}{2 π i} = F (z)$ .

4.1step 1.2step 3.1L2L3givenalgebra

Power series and coefficients. For $∣ z - z_{0} ∣ < r$ the kernel expansion of [step 1.2] is uniformly convergent on $C_{r}$ , so termwise integration of the identity of [step 3.1] gives $F (z) = \sum_{n \geq 0} a_{n} (r) (z - z_{0})^{n}$ with $a_{n} (r) : = \frac{1}{2 π i} \oint_{C_{r}} \frac{F (w)}{(w - z_{0})^{n + 1}} d w$ , and $∥ a_{n} (r) ∥ \leq M (r) / r^{n}$ by [L2]. Two radii $r_{1} < r_{2} < R$ give two power series with the same sum for every real $ζ$ with $∣ ζ ∣ < r_{1}$ after the translation $z = z_{0} + ζ$ ; applying [L3] to these series centered at $0$ gives $a_{n} (r_{1}) = a_{n} (r_{2})$ for every $n$ ; writing $a_{n}$ for the common value, $F$ is represented on $D (z_{0}, R)$ by the norm-convergent series $\sum a_{n} (z - z_{0})^{n}$ , and in particular the coefficient integrals are independent of $r$ .

5.1step 3.1L2givenalgebra∎

Norm- $C^{\infty}$ regularity and Cauchy estimates. Since $∥ a_{n} ∥ \leq M (r^{'}) / r^{' n}$ for every $0 < r^{'} < R$ , for each $0 < ρ < r^{'}$ the differentiated series $\sum_{n \geq 1} n a_{n} (z - z_{0})^{n - 1}$ is dominated on $∣ z - z_{0} ∣ \leq ρ$ by $\sum_{n \geq 1} n M (r^{'}) ρ^{n - 1} / r^{' n} = M (r^{'}) r^{' - 1} (1 - ρ / r^{'})^{- 2} < \infty$ , so it converges uniformly there; the standard difference-quotient estimate $∣ \frac{(z + h - z_{0})^{n} - (z - z_{0})^{n}}{h} - n (z - z_{0})^{n - 1} ∣ \leq \frac{n (n - 1)}{2} ∣ h ∣ \sum_{k} (\binom{n - 2}{k}) ∣ z - z_{0} ∣^{k} ∣ h ∣^{n - 2 - k}$ together with the same geometric majorant shows that the difference quotients of the sum converge to the differentiated sum, so $F$ is complex-differentiable with $F^{'} = \sum n a_{n} (\cdot - z_{0})^{n - 1}$ ; iterating gives $F^{(n)} (z_{0}) = n! a_{n}$ for every $n$ , so $F$ is norm- $C^{\infty}$ and the estimate $∥ F^{(n)} (z_{0}) ∥ = n! ∥ a_{n} ∥ \leq n! M (r) / r^{n}$ follows from $∥ a_{n} ∥ \leq M (r) / r^{n}$ at any $0 < r < R$ . Together with [step 3.1] this proves the formula, the expansion with radius-independent coefficients, and the Cauchy estimates, and no choice principle beyond Countable Choice was used.

Remarks

The circle integral is the norm limit of its Riemann sums: the parametrized integrand is continuous on the compact interval, so its Bochner integral is the limit of the Riemann sums of any sequence of partitions of mesh tending to zero, by uniform continuity and the norm inequality. The proof above separates the three mechanisms usually conflated in the scalar Cauchy theorem: the continuous filled quotient has zero triangle integrals even at its exceptional point, its primitive gives the circle vanishing, and the geometric expansion produces the coefficients; the strict margin $r < R$ keeps every circle compactly contained in the disc of holomorphy.

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Banach-valued power series are determined by their real values

Statement

Let $Y$ be a complex normed vector space (Real and complex scalar conventions for normed spaces), let $z_{0} \in R$ , $r > 0$ , and let $(a_{n})_{n \geq 0}, (b_{n})_{n \geq 0} \subseteq Y$ be such that both series $\sum_{n \geq 0} a_{n} (z - z_{0})^{n}$ and $\sum_{n \geq 0} b_{n} (z - z_{0})^{n}$ converge in $Y$ for every complex $z$ with $∣ z - z_{0} ∣ < r$ (Series and absolute convergence in a normed space). If $\sum_{n \geq 0} a_{n} (t - z_{0})^{n} = \sum_{n \geq 0} b_{n} (t - z_{0})^{n}$ for every real $t$ with $∣ t - z_{0} ∣ < r$ , then $a_{n} = b_{n}$ for every $n$ , and consequently the two sums agree on the whole disc $∣ z - z_{0} ∣ < r$ . No choice principle is used.

Facts & Assumptions

Given: A complex normed vector space $Y$ , a real centre $z_{0}$ , a radius $r > 0$ , sequences $(a_{n})_{n \geq 0}$ and $(b_{n})_{n \geq 0}$ in $Y$ whose series converge on the disc $∣ z - z_{0} ∣ < r$ , the equality of the two sums at every real point $t$ of that disc, and the coefficient differences $d_{n} : = a_{n} - b_{n}$ ; powers are read with the convention $h^{0} = 1$ .

[L1]

A series $\sum_{n = 0}^{\infty} x_{n}$ in a normed space $V$ converges exactly when its partial sums $s_{m} = \sum_{n < m} x_{n}$ converge, and its sum is then $\lim_{m} s_{m}$ ; if $\sum_{n} x_{n}$ and $\sum_{n} y_{n}$ converge, then $\sum_{n} (x_{n} - y_{n})$ converges to the difference of their sums, because its partial sums are the differences of the two partial sums (Series and absolute convergence in a normed space).

[L2]

In a normed space $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ and $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ , and $∥ w ∥ \geq 0$ with $∥ w ∥ = 0$ only for $w = 0$ ; consequently $∣ ∥ u ∥ - ∥ v ∥ ∣ \leq ∥ u - v ∥$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

A complex normed space is a complex vector space with a norm satisfying the same separation and triangle clauses, absolute homogeneity being read with the complex modulus; every estimate that uses only these clauses is valid over either scalar field (Real and complex scalar conventions for normed spaces).

Proof

technique · direct

1.1L1givenalgebra

For every real $h$ with $∣ h ∣ < r$ the series $\sum_{n \geq 0} d_{n} h^{n}$ converges in $Y$ and has sum $0$ : at the point $z = z_{0} + h$ both given series converge, and the partial sums of the difference series are the differences of the corresponding partial sums of the two given series, so they converge to the difference of the two sums, which the hypothesis makes $0$ .

1.2L1L2L3algebra

Continuity at the centre. Let $(c_{j})_{j \geq 0} \subseteq Y$ and $ρ > 0$ be such that $\sum_{j \geq 0} c_{j} h^{j}$ converges for every real $h$ with $∣ h ∣ < ρ$ . Then its sum $S (h)$ satisfies $S (h) \to c_{0}$ as $h \to 0$ . Indeed, put $q : = ρ / 2$ . Convergence at $h = q$ makes the partial sums Cauchy, so their successive differences $c_{j} q^{j}$ tend to $0$ ; a sequence in a normed space that tends to $0$ is bounded, so there is $M < \infty$ with $∥ c_{j} ∥ q^{j} \leq M$ for all $j$ . For $∣ h ∣ \leq q / 2$ and every $N$ the tail bound $∥ \sum_{j > N} c_{j} h^{j} ∥ \leq \sum_{j > N} ∥ c_{j} ∥ ∣ h ∣^{j} \leq M \sum_{j > N} 2^{- j}$ holds: the tail is the limit of its partial sums, the norm is continuous by [L2], and each partial sum is estimated by the triangle inequality. The finite part $\sum_{j \leq N} c_{j} h^{j}$ tends to $c_{0}$ as $h \to 0$ , and $S (0) = c_{0}$ . Hence for $ε > 0$ one chooses $N$ with $M \sum_{j > N} 2^{- j} < ε / 2$ and then $h$ so small that $∥ \sum_{j \leq N} c_{j} h^{j} - c_{0} ∥ < ε / 2$ , giving $∥ S (h) - c_{0} ∥ < ε$ .

2.1step 1.1step 1.2L1algebra

For every $n \geq 0$ : if $d_{0} = \dots = d_{n - 1} = 0$ , then $d_{n} = 0$ . Indeed, the series $T_{n} (h) : = \sum_{j \geq 0} d_{n + j} h^{j}$ converges at $h = 0$ with sum $d_{n}$ , and for $0 < ∣ h ∣ < r$ the vanishing of the initial coefficients makes the partial sums of $\sum_{k \geq 0} d_{k} h^{k}$ equal to $h^{n}$ times the partial sums of $T_{n} (h)$ , so that $\sum_{k \geq 0} d_{k} h^{k} = h^{n} T_{n} (h)$ ; by [step 1.1] the left side converges to $0$ , hence $T_{n} (h) = 0$ for $0 < ∣ h ∣ < r$ and the series $T_{n} (h)$ converges for every real $∣ h ∣ < r$ . Applying [step 1.2] with $c_{j} : = d_{n + j}$ and any $ρ \in (0, r)$ gives $d_{n} = T_{n} (0) = \lim_{h \to 0} T_{n} (h) = 0$ .

3.1step 2.1given

Induction on $n$ : [step 2.1] says that the vanishing of $d_{0}, \dots, d_{n - 1}$ forces the vanishing of $d_{n}$ for every $n$ , so the set of indices with $d_{n} = 0$ contains $0$ and is closed under successors; it is therefore all of $N$ . Hence $a_{n} = b_{n}$ for every $n$ .

4.1step 3.1given∎

For every complex $z$ with $∣ z - z_{0} ∣ < r$ the two series are termwise identical, hence, both being convergent there, they have the same sum; this proves the agreement on the whole disc, and the argument used only limits, norm estimates and induction, so no choice principle was used.

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Taylor expansion with integral remainder for Banach-valued curves

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited integral and semigroup suppliers.

Let $Y$ be a Banach space over $K \in {R, C}$ , let $I \subseteq R$ be an interval, let $n \geq 0$ be an integer, and let $u : I \to Y$ . For a nondegenerate interval, $u \in C^{n + 1} (I; Y)$ means that $u$ is continuous on $I$ , its restriction to $int I$ has norm-continuous derivatives through order $n + 1$ , and each derivative extends continuously to $I$ . Derivatives on the interior are taken with respect to the real parameter, using the underlying real Banach space when $Y$ is complex (Fréchet derivative between Banach spaces), and $u^{(k)}$ denotes the continuous extension at any included endpoint; set $u^{(0)} = u$ . Assume this regularity. Then for all $t \in I$ and $h \in R$ with $[\min {t, t + h}, \max {t, t + h}] \subseteq I$ , $u (t + h) = \sum_{k = 0}^{n} u^{(k)} (t) \frac{h^{k}}{k!} + \frac{1}{n!} \int_{t}^{t + h} (t + h - s)^{n} u^{(n + 1)} (s) d s,$ the integral being a Bochner integral (Bochner-integrable function); for $h < 0$ the symbol $\int_{t}^{t + h}$ denotes the oriented Bochner interval integral $- \int_{t + h}^{t}$ . For $n = 0$ the formula is the fundamental theorem of calculus. For $h = 0$ the formula is understood as $u (t) = u (t)$ ; this also covers singleton intervals without assigning higher derivatives there. No choice principle beyond Countable Choice is used.

Facts & Assumptions

Given: Countable Choice; a Banach space $Y$ over $K$ , an interval $I \subseteq R$ , an integer $n \geq 0$ , a curve $u : I \to Y$ continuous on $I$ whose real-parameter derivatives through order $n + 1$ on $int I$ extend continuously to $I$ when $I$ is nondegenerate, and points $t \in I$ , $h \in R$ with $[\min {t, t + h}, \max {t, t + h}] \subseteq I$ . Write $u^{(k)}$ for these extensions and $u^{(0)} = u$ ; for $h < 0$ use $\int_{t}^{t + h} f : = - \int_{t + h}^{t} f$ , and for $h = 0$ read the formula as $u (t) = u (t)$ , including singleton $I$ .

[L1]

A continuous $f : [a, b] \to Y$ is Bochner integrable; its primitive $G (t) = \int_{a}^{t} f$ is differentiable with $G^{'} = f$ , and for a continuous curve $φ$ of class $C^{1}$ on $(a, b)$ whose derivative extends continuously to $[a, b]$ one has $\int_{a}^{b} φ^{'} = φ (b) - φ (a)$ (Fundamental theorem of calculus for Banach-valued continuous curves).

[L2]

The Bochner integral is linear in its integrand, so $\int (α f + β g) = α \int f + β \int g$ on a fixed interval, and $∥ \int_{E} f ∥ \leq \int_{E} ∥ f ∥$ (Linearity of the Bochner integral, Bochner integral norm inequality). Scalar and vector operations are continuous: $∥ λ v - λ_{0} v_{0} ∥ \leq ∣ λ - λ_{0} ∣ ∥ v ∥ + ∣ λ_{0} ∣ ∥ v - v_{0} ∥$ .

Proof

technique · direct

1.1L1givenalgebra

If $h = 0$ , the formula is $u (t) = u (t)$ for every $n$ , including singleton $I$ . Henceforth let $h \neq 0$ , so $I$ is nondegenerate; its interior is dense in $I$ , making each continuous derivative extension unique. Base case $n = 0$ : for $h > 0$ , the continuous curve $u$ is differentiable inside $[t, t + h]$ with derivative extending continuously there as $u^{'}$ , so [L1] gives $\int_{t}^{t + h} u^{'} (s) d s = u (t + h) - u (t)$ ; for $h < 0$ , [L1] on $[t + h, t]$ and the oriented convention give $\int_{t}^{t + h} u^{'} = - \int_{t + h}^{t} u^{'} = - (u (t) - u (t + h)) = u (t + h) - u (t)$ .

1.2L1L2givenalgebra

Integration by parts identity. For $1 \leq k \leq n$ put $φ (s) : = \frac{(t + h - s)^{k}}{k!} u^{(k)} (s)$ ; on the interior of the ordered segment the scalar-times-vector product rule gives $φ^{'} (s) = - \frac{(t + h - s)^{k - 1}}{(k - 1)!} u^{(k)} (s) + \frac{(t + h - s)^{k}}{k!} u^{(k + 1)} (s)$ , because $φ (s + Δ) - φ (s) = (α (s + Δ) - α (s)) v (s + Δ) + α (s) (v (s + Δ) - v (s))$ for the scalar $α (s) = (t + h - s)^{k} / k!$ and the vector $v = u^{(k)}$ , and scalar multiplication is continuous by [L2]. For $h > 0$ , $φ$ is continuous on $[t, t + h]$ with $φ^{'}$ extending continuously there (the derivatives of $u$ up to order $k + 1$ have continuous extensions), so [L1] gives $\int_{t}^{t + h} φ^{'} = φ (t + h) - φ (t) = - \frac{h^{k}}{k!} u^{(k)} (t)$ ; rearranging with the linearity [L2] yields $\frac{1}{(k - 1)!} \int_{t}^{t + h} (t + h - s)^{k - 1} u^{(k)} (s) d s = \frac{h^{k}}{k!} u^{(k)} (t) + \frac{1}{k!} \int_{t}^{t + h} (t + h - s)^{k} u^{(k + 1)} (s) d s$ . For $h < 0$ the same computation is applied on the interval $[t + h, t]$ with the oriented sign, and the displayed identity is unchanged because both integrals acquire one sign reversal.

2.1step 1.2L1givenalgebra

Induction step. Assume the formula holds with $n - 1 \geq 0$ in place of $n$ for every curve of class $C^{n}$ ; applying it to the $C^{n + 1}$ curve $u$ gives $u (t + h) = \sum_{k = 0}^{n - 1} u^{(k)} (t) \frac{h^{k}}{k!} + \frac{1}{(n - 1)!} \int_{t}^{t + h} (t + h - s)^{n - 1} u^{(n)} (s) d s$ , and [step 1.2] with $k = n$ rewrites the last term as $\frac{h^{n}}{n!} u^{(n)} (t) + \frac{1}{n!} \int_{t}^{t + h} (t + h - s)^{n} u^{(n + 1)} (s) d s$ ; substituting gives the formula with $n$ , all integrands being continuous hence Bochner integrable on the compact interval by [L1].

3.1step 1.1step 2.1given∎

Conclusion. [step 1.1] is the case $n = 0$ for both signs of $h$ and [step 2.1] carries the induction from $n - 1$ to $n$ for every $n \geq 1$ , so the formula holds for all $n \geq 0$ ; the proof used only the one-dimensional fundamental theorem, the product rule for a scalar and a vector curve, and linearity of the Bochner integral, hence no choice principle beyond Countable Choice was used.

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The generator of the contour semigroup is the sectorial 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 $A$ be a sectorial operator of angle $δ \in (0, π / 2]$ with vertex $0$ on a complex Banach space $X$ (Sectorial operator with the semigroup sign convention) and let $(T (z))_{z \in Σ_{δ} \cup {0}}$ be the contour family of The Dunford contour integral defines a bounded holomorphic family on the sector, shown in The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex to be a bounded analytic semigroup of angle $δ$ . Then the generator of the strongly continuous semigroup $(T (t))_{t \geq 0}$ (Infinitesimal generator of a C0-semigroup) is $A$ , and $(T (t))_{t \geq 0}$ is the unique strongly continuous semigroup generated by $A$ within the class of exponentially bounded semigroups. No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ on the Banach space $X$ , its contour semigroup $(T (t))_{t \geq 0}$ , the generator $B$ of that semigroup, a real $ω$ with $∥ T (t) ∥ \leq M e^{ω t}$ for all $t \geq 0$ , and a fixed real $λ > \max {ω, 0}$ .

[L1]

For the contour semigroup, $T (t) X \subseteq D (A)$ , $A T (t) = T^{'} (t)$ , $t \mapsto T (t)$ is norm- $C^{1}$ for $t > 0$ , and $T^{'} (t) = \frac{1}{2 π i} \int_{Γ} λ e^{λ t} R (λ, A) d λ$ (Smoothing estimates for the semigroup generated by a sectorial operator, The Dunford contour integral defines a bounded holomorphic family on the sector).

[L2]

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

[L3]

The fundamental theorem of calculus for Banach-valued curves, and average convergence: for continuous $g$ , $h^{- 1} \int_{0}^{h} g (s) d s \to g (0)$ (Fundamental theorem of calculus for Banach-valued continuous curves, Average convergence for a continuous Banach-valued function).

[L4]

Sectoriality with vertex $0$ makes $A$ closed and densely defined and puts every positive real $λ$ in $ρ (A)$ (Sectorial operator with the semigroup sign convention).

[L5]

For any strongly continuous semigroup with generator $C$ , if $y \in D (C)$ then its orbit remains in $D (C)$ and $C S (t) y = S (t) C y$ (The generator commutes with the semigroup on its domain).

[L6]

For $λ \in ρ (A)$ , $R (λ, A) X = D (A)$ , $A R (λ, A) = λ R (λ, A) - I$ , and $R (λ, A) (λ I - A) y = y$ for $y \in D (A)$ (Resolvent and spectrum of a closed operator on a Banach space).

[L7]

The generator $B$ is defined by $x \in D (B)$ exactly when $h^{- 1} (T (h) x - x)$ has a limit as $h ↓ 0$ , and that limit is $B x$ (Infinitesimal generator of a C0-semigroup).

Proof

technique · direct

1.1L1L3L6L7givenalgebra

The inclusion $A \subseteq B$ . First let $y \in D (A)$ and $s > 0$ . The contour formula for $T (s) A y$ and [L1] give $T (s) A y = \frac{1}{2 π i} \int_{Γ} e^{λ s} R (λ, A) A y d λ = \frac{1}{2 π i} \int_{Γ} e^{λ s} (λ R (λ, A) y - y) d λ = T^{'} (s) y = A T (s) y .$ Here $R (λ, A) A y = λ R (λ, A) y - y$ follows from [L6], and $(2 π i)^{- 1} \int_{Γ} e^{λ s} d λ = 0$ : close a truncated keyhole, use the entire primitive $e^{λ s} / s$ , and let its exponentially decaying outer arc tend to zero. Now for $x \in D (A)$ and $h > 0$ , the fundamental theorem [L3] on $[ε, h]$ , followed by $ε ↓ 0$ using strong continuity at $0$ , gives $T (h) x - x = \int_{0}^{h} T (s) A x d s$ . Dividing by $h$ and applying average convergence [L3] yields $h^{- 1} (T (h) x - x) \to A x$ . By the generator definition in [L7], $x \in D (B)$ and $B x = A x$ .

1.2L1L3L4givenalgebra

The resolvent computation. Fix $x \in X$ and $λ > \max {ω, 0}$ , and put $v_{η, R} : = \int_{η}^{R} e^{- λ t} T (t) x d t$ for $0 < η < R$ . On the compact interval $[η, R]$ , both $t \mapsto T (t) x$ and $t \mapsto A T (t) x = T^{'} (t) x$ are continuous, so the orbit is continuous in the graph norm of the closed operator $A$ . Its graph-norm Bochner integral therefore lies in $D (A)$ and satisfies $A v_{η, R} = \int_{η}^{R} e^{- λ t} A T (t) x d t$ . Since $(d / d t) [e^{- λ t} T (t) x] = e^{- λ t} (A T (t) x - λ T (t) x)$ , the fundamental theorem [L3] gives $(λ I - A) v_{η, R} = \int_{η}^{R} e^{- λ t} (λ T (t) x - A T (t) x) d t = - \int_{η}^{R} \frac{d}{d t} (e^{- λ t} T (t) x) d t = e^{- λ η} T (η) x - e^{- λ R} T (R) x$ . As $η ↓ 0$ and $R \to \infty$ , $v_{η, R} \to v : = \int_{0}^{\infty} e^{- λ t} T (t) x d t$ by the exponential bound and strong continuity at $0$ , while the displayed right side tends to $x$ because $λ > 0$ . Closedness of $A$ now gives $v \in D (A)$ and $(λ I - A) v = x$ ; since $λ > 0$ and $A$ is sectorial, $λ \in ρ (A)$ and $v = R (λ, A) x$ .

2.1step 1.2L2L4L6givenalgebra

The resolvents agree and $B = A$ . By [step 1.2] the vector $v = \int_{0}^{\infty} e^{- λ t} T (t) x d t$ equals $R (λ, A) x$ ; by the Laplace formula [L2] the same integral equals $R (λ, B) x$ . Hence $R (λ, B) x = R (λ, A) x$ for every $x \in X$ . Their ranges agree and equal $D (B) = D (A)$ ; for each $y$ in this common domain, applying the inverses gives $(λ I - B) y = (λ I - A) y$ , so $B = A$ .

2.2step 1.1L3L4L5givenalgebra

Uniqueness among exponentially bounded semigroups. Let $S$ and $T$ be two strongly continuous semigroups with generator $A$ and exponential bounds, fix $t > 0$ and $x \in D (A)$ , and set $g (s) : = S (t - s) T (s) x$ for $s \in [0, t]$ . For $0 < s < t$ , write the difference quotient as $\frac{g (s + h) - g (s)}{h} = \frac{S (t - s - h) - S (t - s)}{h} T (s) x + S (t - s - h) \frac{T (s + h) x - T (s) x}{h} .$ By [L5], $T (s) x \in D (A)$ , the first term tends to $- S (t - s) A T (s) x$ , and the second tends to $S (t - s) A T (s) x$ ; hence $g^{'} (s) = 0$ . Continuity at the endpoints and the fundamental theorem [L3] show that $g$ is constant, so $S (t) x = T (t) x$ . Since $D (A)$ is dense by [L4] and $S (t), T (t)$ are bounded, this extends to all $x \in X$ .

3.1step 2.1step 2.2given∎

Assembly. [step 2.1] identifies the generator of the contour semigroup with $A$ , and [step 2.2] proves uniqueness in the exponentially bounded class; no choice principle beyond Dependent Choice was used, since only the contour construction, the Laplace representation and the fundamental theorem were invoked.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Complex sector and bounded analytic semigroup

Definition

For $δ \in (0, π]$ put $Σ_{δ} : = {z \in C ∖ {0} : ∣ \arg z ∣ < δ},$ the open sector of half-angle $δ$ around the positive real axis (A complex domain is a nonempty connected open subset of $C$ ; we use the principal argument in $(- π, π]$ ). Let $X$ be a Banach space over $C$ (Banach space). A family $(T (z))_{z \in Σ_{δ} \cup {0}} \subseteq B (X)$ (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators) is an analytic semigroup of angle $δ \in (0, π / 2]$ if:

(i) $T (0) = I$ and $T (z_{1} + z_{2}) = T (z_{1}) T (z_{2})$ for all $z_{1}, z_{2} \in Σ_{δ}$ ;

(ii) $z \mapsto T (z)$ is holomorphic on $Σ_{δ}$ in the operator norm: the difference quotients $h^{- 1} (T (z + h) - T (z))$ converge in $B (X)$ for every $z \in Σ_{δ}$ ;

(iii) $\lim_{Σ_{δ^{'}} ∋ z \to 0} T (z) x = x$ for every $x \in X$ and every $0 < δ^{'} < δ$ .

It is a bounded analytic semigroup of angle $δ$ if in addition

(iv) $\sup_{z \in Σ_{δ^{'}}} ∥ T (z) ∥ < \infty$ for every $0 < δ^{'} < δ$ .

Its generator is the infinitesimal generator of the strongly continuous semigroup $(T (t))_{t \geq 0}$ (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup), and its angle is the supremum of the $δ$ for which such a family exists and extends the given one.

Strong continuity is required only at the vertex and only in the strong operator topology; holomorphy is asserted on the open sector, not at $0$ . For a real Banach space $X$ the definition is applied through a complexification (Canonical Banach complexification of a real Banach space, Real and complex scalar conventions for normed spaces).

Remarks

The sector $Σ_{δ} \cup {0}$ is a convex cone for $δ \leq π / 2$ , so $z_{1} + z_{2}$ stays in the index set in (i); the vertex is the only boundary point at which values are prescribed. Condition (iii) is an assumption on the approach to the vertex along every strictly smaller sector, and it implies that $(T (t))_{t \geq 0}$ is a strongly continuous semigroup, since $[0, \infty) \subseteq Σ_{δ} \cup {0}$ .
No norm continuity at $0$ is asserted: when the generator is unbounded the family is only strongly continuous there, as the companion counterexample records. Likewise $T$ is not required to be holomorphic at $0$ ; only the values $T (z)$ for $z$ in the open sector carry the holomorphy of (ii).
Condition (iv) is a boundedness requirement on every strictly smaller sector and not on all of $Σ_{δ}$ ; this is the distinction between a bounded analytic semigroup and an analytic semigroup whose norm may blow up near the boundary of the sector.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Sectorial operator with the semigroup sign convention

Definition

Let $X$ be a Banach space over $K \in {R, C}$ (Banach space, Real and complex scalar conventions for normed spaces) and let $A : D (A) \subseteq X \to X$ be a closed densely defined linear operator with resolvent $R (λ, A) = (λ I - A)^{- 1}$ (Resolvent and spectrum of a closed operator on a Banach space, Densely defined, closed and closable operators, and cores, Unbounded linear operators: domain, graph and extension).

For a real $X$ , all complex resolvents below are those of the closed complexified operator $A_{C} (x, y) = (A x, A y)$ on $D (A) \times D (A)$ in the canonical complexification (Canonical Banach complexification of a real Banach space). Closedness and density follow coordinatewise, since its norm is equivalent to the product norm.

For $δ \in (0, π / 2]$ and $ω \in R$ , $A$ (or the pair $(A, ω)$ ) is sectorial of angle $δ$ with vertex $ω$ in the $e^{t A}$ convention if the open sector $ω + Σ_{π / 2 + δ}$ (Complex sector and bounded analytic semigroup) is contained in the resolvent set $ρ (A)$ , and for every $ε \in (0, δ)$ there is a constant $M_{ε} \geq 1$ with $∥ R (λ, A) ∥ \leq \frac{M_{ε}}{∣ λ - ω ∣} for all λ \in ω + Σ_{π / 2 + δ - ε} .$

The sign dictionary

The definition is equivalent to the pair of statements that the spectrum of $A$ is contained in the complementary closed left sector $σ (A) \subseteq ω + ({λ \neq 0 : ∣ \arg (- λ) ∣ \leq π / 2 - δ} \cup {0})$ and that the stated $M_{ε} / ∣ λ - ω ∣$ bound holds on the right-opening sector. Writing $B : = - A + ω I$ one has $(μ I - B) = (μ - ω) I + A = - ((ω - μ) I - A),$ hence $R (μ, B) = - R (ω - μ, A)$ ; substituting $λ = ω - μ$ , the resolvent bound of $A$ on $ω + Σ_{π / 2 + δ - ε}$ becomes the bound $∥ R (μ, B) ∥ \leq M_{ε} / ∣ μ ∣$ on the reflected left-opening sector $- Σ_{π / 2 + δ - ε}$ for $μ$ , so that sector lies in $ρ (B)$ and the spectrum of $B$ lies in the closed sector ${μ \neq 0 : ∣ \arg μ ∣ \leq π / 2 - δ} \cup {0}$ .

This dictionary is why every theorem on this page states its convention: a source that calls $A$ "sectorial" for the opposite operator $- A$ , or that writes $e^{- t (- A)}$ , is using the Pazy-Lunardi sign and its sector and angle must be reflected before transfer. The free use of laplace-transform-shaped formulas below is always in the $e^{t A}$ convention fixed here: the resolvent sector of $A$ opens around the positive real direction, the spectral sector lies to the left, and positive time corresponds to an integral of $e^{λ z} R (λ, A)$ .

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The Dunford contour integral defines a bounded holomorphic family on the sector

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 $A$ be sectorial of angle $δ \in (0, π / 2]$ with vertex $0$ on a complex Banach space $X$ (Sectorial operator with the semigroup sign convention, Banach space). For $r > 0$ and $θ \in (π / 2, π / 2 + δ)$ let $Γ (r, θ)$ consist of the lower ray $s e^{- i θ}$ for $s \geq r$ , the circular arc $r e^{i α}$ for $- θ \leq α \leq θ$ , and the upper ray $s e^{i θ}$ for $s \geq r$ , oriented counterclockwise around the spectrum. For $R > r$ let $Γ_{R} (r, θ)$ be the corresponding truncated path. Its integrand $λ \mapsto e^{λ z} R (λ, A)$ is $B (X)$ -valued, and its contour integral is the Bochner integral in the Banach space $B (X)$ with the operator norm; this space is Banach because $X$ is Banach (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Linearity of the Bochner integral, Bochner integral norm inequality, If (Y) is Banach then (\mathcal B(X,Y)) is Banach). For $z \in Σ_{δ}$ (Complex sector and bounded analytic semigroup) choose an admissible angle satisfying $π / 2 + ∣ \arg z ∣ < θ < π / 2 + δ$ and set $T (z) : = \frac{1}{2 π i} \lim_{R \to \infty} \int_{Γ_{R} (r, θ)} e^{λ z} R (λ, A) d λ,$ where the limit is taken in the complete operator-norm space $B (X)$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, If (Y) is Banach then (\mathcal B(X,Y)) is Banach). Then:

the integral converges absolutely for every such $z$ ; for each compact $K \subset Σ_{δ}$ one admissible angle can be chosen for all $z \in K$ , and the truncated integrals converge absolutely and locally uniformly in operator norm on $K$ ;
the value is independent of $r > 0$ and of the admissible angle $θ$ ;
$z \mapsto T (z)$ is norm-holomorphic and $T^{'} (z) = \frac{1}{2 π i} \int_{Γ (r, θ)} λ e^{λ z} R (λ, A) d λ;$
$\sup_{z \in Σ_{δ^{'}}} ∥ T (z) ∥ < \infty$ for every $δ^{'} < δ$ .

No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with vertex $0$ on a complex Banach space $X$ , with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ on $∣ \arg λ ∣ < π / 2 + δ - ε$ for every $ε \in (0, δ)$ (Sectorial operator with the semigroup sign convention); fixed $r > 0$ ; a compact $K \subset Σ_{δ}$ with $δ_{K} : = \max_{z \in K} ∣ \arg z ∣ < δ$ , $m_{K} : = \min_{z \in K} ∣ z ∣ > 0$ , $M_{K} : = \max_{z \in K} ∣ z ∣$ ; a fixed $θ$ with $π / 2 + δ_{K} < θ < π / 2 + δ$ and $ε > 0$ with $θ < π / 2 + δ - ε$ ; the truncated contours $Γ_{R} (r, θ)$ ; and $ψ_{z} (λ) : = e^{λ z} R (λ, A)$ .

[L1]

$∣ \arg λ ∣ \leq θ$ implies $λ \in ρ (A)$ and $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ with $M_{ε}$ as in the givens (Sectorial operator with the semigroup sign convention).

[L2]

$λ \mapsto R (λ, A)$ is norm-holomorphic on $ρ (A)$ with derivative $- R (λ, A)^{2}$ , and for every $z \in C$ the maps $λ \mapsto e^{λ z} R (λ, A)$ and $λ \mapsto λ e^{λ z} R (λ, A)$ are norm-holomorphic on $ρ (A)$ (Resolvent identity and holomorphy for a closed operator).

[L3]

The open sector $S_{ε} : = {λ \neq 0 : ∣ \arg λ ∣ < π / 2 + δ - ε}$ has half-angle $< π$ and omits the negative real axis, so it is star-shaped with base point any positive real number: a segment from a positive real number to a point of the sector cannot contain $0$ and its arguments stay in the convex cone spanned by the positive axis and the endpoint; every $B (X)$ -valued function continuous and complex-differentiable on a star-shaped domain has vanishing integral over closed piecewise $C^{1}$ contours in it, since $B (X)$ is Banach by [L5] (Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains).

[L4]

For the curve integral of a continuous integrand, $∥ \int_{γ} f d λ ∥ \leq \int_{γ} ∥ f ∥ ∣ d λ ∣$ and the integral is linear in $f$ (Bochner integral norm inequality, Linearity of the Bochner integral).

[L5]

Since $X$ is Banach, $B (X) = B (X, X)$ is Banach in the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach).

Proof

technique · direct

1.1givenalgebra

Angle geometry on $K$ . For $z \in K$ , the upper-ray angle satisfies $π / 2 < θ + \arg z < 3 π / 2$ , since $θ + \arg z \geq θ - δ_{K} > π / 2$ and $θ + \arg z \leq θ + δ_{K} < π / 2 + δ + δ_{K} < 3 π / 2$ . The lower-ray angle satisfies $- 3 π / 2 < - θ + \arg z < - π / 2$ , since $- θ + \arg z \leq - θ + δ_{K} < - π / 2$ and $- θ + \arg z \geq - θ - δ_{K} > - π / 2 - δ - δ_{K} > - 3 π / 2$ . Thus $\cos (\pm θ + \arg z) < 0$ for both rays. The continuous function $z \mapsto - \cos (\pm θ + \arg z)$ is positive on $K$ , so $c_{K} : = \min_{z \in K, θ^{'} = \pm θ} (- \cos (θ^{'} + \arg z)) > 0$ , and for $λ = s e^{\pm i θ}$ one has $Re (λ z) = s ∣ z ∣ \cos (\pm θ + \arg z) \leq - c_{K} s ∣ z ∣$ .

1.2L1L2L3L5given

The integrand is holomorphic on a star-shaped sector. By [L1] the sector $S_{ε}$ of [L3] lies in $ρ (A)$ , and by [L2] both $λ \mapsto e^{λ z} R (λ, A)$ and $λ \mapsto λ e^{λ z} R (λ, A)$ are norm-holomorphic on it; $S_{ε}$ is star-shaped with base point any positive real by [L3]. Since $B (X)$ is Banach by [L5], the Banach-valued Cauchy theorem applies to these $B (X)$ -valued maps.

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Absolute convergence and local uniformity. For $z \in K$ and $λ = s e^{\pm i θ}$ on the rays, [step 1.1] and [L1] give $∥ ψ_{z} (λ) ∥ \leq e^{- c_{K} s ∣ z ∣} M_{ε} / s \leq e^{- c_{K} m_{K} s} M_{ε} / s$ , which is integrable over $s \geq r$ ; on the arc $∣ λ ∣ = r$ one has $∥ ψ_{z} (λ) ∥ \leq e^{M_{K} r} M_{ε} / r$ , an integrable bound on a compact interval. Hence $\int_{Γ} ∥ ψ_{z} (λ) ∥ ∣ d λ ∣ < \infty$ and, by [L4], $∥ \int_{Γ_{R}} ψ_{z} d λ - \int_{Γ_{R^{'}}} ψ_{z} d λ ∥ \leq 2 M_{ε} \int_{R}^{R^{'}} e^{- c_{K} m_{K} s} d s / s$ for $R < R^{'}$ , a bound independent of $z \in K$ tending to $0$ . Thus the truncated integrals form a Cauchy family in $B (X)$ and converge there by [L5]; this is the operator-norm limit, uniformly on $K$ , and the integrals converge absolutely.

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Independence of the inner radius. Fix $θ$ and $0 < r_{1} < r_{2}$ and $R > r_{2}$ . The truncated paths $Γ_{R} (r_{1}, θ)$ and $Γ_{R} (r_{2}, θ)$ have the same initial point $R e^{- i θ}$ and the same terminal point $R e^{i θ}$ , so their concatenation with the reversal of the second is a closed piecewise $C^{1}$ contour lying in the star-shaped sector $S_{ε}$ of [step 1.2], where $ψ_{z}$ is holomorphic; by the Banach-valued Cauchy theorem in $B (X)$ , applicable by [L5], [L3] its integral vanishes, hence $\int_{Γ_{R} (r_{1}, θ)} ψ_{z} d λ = \int_{Γ_{R} (r_{2}, θ)} ψ_{z} d λ$ for every $R > r_{2}$ ; letting $R \to \infty$ and using [step 2.1] gives equality of the limits.

3.2step 1.2step 2.1L3L5givenalgebra

Independence of the angle. Fix $r$ and $π / 2 < θ_{1} < θ_{2} < π / 2 + δ$ , both admissible for the given $z$ , and $R > r$ . Let $A_{R}^{+}$ be the counterclockwise arc $R e^{i α}$ , $α \in [θ_{1}, θ_{2}]$ , and $A_{R}^{-}$ its reflection $α \in [- θ_{2}, - θ_{1}]$ ; then $Γ_{R} (r, θ_{1}) + A_{R}^{+} - Γ_{R} (r, θ_{2}) + A_{R}^{-}$ is a closed piecewise $C^{1}$ contour in the star-shaped sector $S_{ε}$ (for $θ_{2} < π / 2 + δ - ε$ ), so the Banach-valued Cauchy theorem in $B (X)$ applies by [L5] and [L3] its integral vanishes; hence $\int_{Γ_{R} (r, θ_{1})} ψ_{z} - \int_{Γ_{R} (r, θ_{2})} ψ_{z} = - \int_{A_{R}^{+}} ψ_{z} - \int_{A_{R}^{-}} ψ_{z}$ . On the arcs $∣ λ ∣ = R$ with angle $α$ between $θ_{1}, θ_{2}$ and their reflections one has $∣ α + \arg z ∣ > π / 2$ uniformly, so $∥ ψ_{z} ∥ \leq e^{- c R}$ for a constant $c > 0$ and the right-hand side tends to $0$ as $R \to \infty$ ; hence the two limits agree.

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Norm-holomorphy and the derivative formula. Fix $K$ , the angle $θ$ of [step 1.1] and $r > 0$ . Put $d_{K} : = dist (K, C ∖ Σ_{δ}) > 0$ and choose $h_{0} < \min {c_{K} m_{K}, d_{K}}$ . Then $z + h \in Σ_{δ}$ for $z \in K$ , $∣ h ∣ < h_{0}$ , and the exponential estimate $∣ h^{- 1} (e^{λ h} - 1) - λ ∣ \leq ∣ h ∣ ∣ λ ∣^{2} e^{∣ h ∣ ∣ λ ∣}$ together with [L1] gives, on the rays, $∥ h^{- 1} (ψ_{z + h} (λ) - ψ_{z} (λ)) - λ ψ_{z} (λ) ∥ \leq ∣ h ∣ M_{ε} ∣ λ ∣ e^{- c_{K} ∣ λ ∣ ∣ z ∣ + h_{0} ∣ λ ∣} \leq ∣ h ∣ M_{ε} s e^{- (c_{K} m_{K} - h_{0}) s}$ with $c_{K} m_{K} - h_{0} > 0$ , an integrable bound whose integral tends to $0$ with $h$ , while on the compact arc the same estimate is bounded uniformly and contributes $O (∣ h ∣)$ ; hence for each fixed $R$ the difference quotients of $\int_{Γ_{R}} ψ$ tend to $\int_{Γ_{R}} λ ψ_{z} d λ$ with an error bounded uniformly in $R$ , and letting $R \to \infty$ with the majorant of [step 2.1] and [L4] gives $∥ h^{- 1} (T (z + h) - T (z)) - (2 π i)^{- 1} \int_{Γ} λ e^{λ z} R (λ, A) d λ ∥ \to 0$ ; completeness [L5] ensures that this derivative integral and the limit lie in $B (X)$ , so $T$ is complex-differentiable with that derivative (uniformly on $K$ ).

4.1step 3.1L1L4givenalgebra

Uniform boundedness on smaller sectors. Fix $δ^{'} < δ$ and an angle $θ$ with $π / 2 + δ^{'} < θ < π / 2 + δ$ ; for $z \in Σ_{δ^{'}}$ the value $T (z)$ is, by the independence of the inner radius [step 3.1], computable with $r = 1 / ∣ z ∣$ , so on the rays $Re (λ z) \leq - c s ∣ z ∣$ with $c = c (θ, δ^{'}) > 0$ and the ray contribution is at most $M_{ε} \int_{1}^{\infty} e^{- c σ} d σ / σ$ , while the arc has length at most $2 θ / ∣ z ∣$ , radius $1 / ∣ z ∣$ and integrand norm at most $e M_{ε} ∣ z ∣$ , contributing at most $2 θ e M_{ε}$ ; both bounds are independent of $z$ , so $\sup_{Σ_{δ^{'}}} ∥ T ∥ < \infty$ .

5.1step 2.1step 3.1step 3.2step 3.3step 4.1given∎

Conclusion. [step 2.1] proves claim 1, [step 3.1] and [step 3.2] prove claim 2, [step 3.3] proves claim 3, and [step 4.1] proves claim 4; the argument used only the sectorial resolvent bound, the resolvent holomorphy, the contour integral over closed curves in the star-shaped sector and norm estimates, hence no choice principle beyond Dependent Choice was used.

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The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex

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.

In the setting of The Dunford contour integral defines a bounded holomorphic family on the sector, extend $T$ by $T (0) : = I$ . Then:

$T (z_{1} + z_{2}) = T (z_{1}) T (z_{2})$ for all $z_{1}, z_{2} \in Σ_{δ}$ ;
$T (z) x \to x$ as $Σ_{δ^{'}} ∋ z \to 0$ for every $x \in X$ and every $δ^{'} < δ$ , with the quantitative estimate $∥ T (z) x - x ∥ \leq C_{δ^{'}, r} ∣ z ∣ ∥ A x ∥$ for $x \in D (A)$ and $∣ z ∣ \leq r$ ;

hence $(T (z))_{z \in Σ_{δ} \cup {0}}$ is a bounded analytic semigroup of angle $δ$ in the sense of Complex sector and bounded analytic semigroup. No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with vertex $0$ on a complex Banach space $X$ and the contour family $T (z)$ of The Dunford contour integral defines a bounded holomorphic family on the sector, with $T (0) : = I$ ; fixed $z_{1}, z_{2} \in Σ_{δ}$ and $δ_{0} < δ$ with $z_{1}, z_{2}, z_{1} + z_{2} \in Σ_{δ_{0}}$ ; and for the strong-continuity part a fixed $x \in D (A)$ .

[L1]

$T$ is well defined, norm-holomorphic on $Σ_{δ}$ , independent of the inner radius and of the admissible angle, and $\sup_{z \in Σ_{δ^{'}}} ∥ T (z) ∥ < \infty$ for every $δ^{'} < δ$ (The Dunford contour integral defines a bounded holomorphic family on the sector).

[L2]

For $λ, μ \in ρ (A)$ , $R (λ, A) - R (μ, A) = (μ - λ) R (λ, A) R (μ, A)$ (Resolvent identity and holomorphy for a closed operator). For $μ \in ρ (A) ∖ {0}$ and $x \in D (A)$ , the inverse relation $R (μ, A) (μ x - A x) = x$ gives $R (μ, A) x - μ^{- 1} x = μ^{- 1} R (μ, A) A x$ (Resolvent and spectrum of a closed operator on a Banach space).

[L3]

On any open star-shaped domain $U \subseteq C$ , the integral of a holomorphic scalar- or Banach-valued function over a closed piecewise $C^{1}$ contour in $U$ is zero (Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains). In particular this applies to contours in the sector $S : = {λ \neq 0 : ∣ \arg λ ∣ < π / 2 + δ}$ ; the improper keyhole integrals defining $T$ converge absolutely (The Dunford contour integral defines a bounded holomorphic family on the sector).

[L4]

For the curve integral, $∥ \int_{γ} f ∥ \leq \int_{γ} ∥ f ∥$ and the integral is linear (Bochner integral norm inequality, Linearity of the Bochner integral).

[L5]

The index is the integral $(2 π i)^{- 1} \oint (λ - p)^{- 1} d λ$ , is constant on each connected component off the trace, and vanishes on the unbounded component (The winding number of a closed contour about a point off its trace, The winding number is constant on each connected component of the complement of the trace, The winding number vanishes on the unbounded component of the complement of the trace).

Proof

technique · direct

1.1L1L4givenalgebra

The double integral. Fix admissible contours $Γ_{1} = Γ (r_{1}, θ_{1})$ and $Γ_{2} = Γ (r_{2}, θ_{2})$ with $π / 2 + δ_{0} < θ_{1} < θ_{2} < π / 2 + δ$ and $r_{2} < r_{1}$ ; then every point of $Γ_{1}$ lies outside the interior of $Γ_{2}$ and every point of $Γ_{2}$ lies inside the interior of $Γ_{1}$ . By [L1] and [L4] the product $T (z_{1}) T (z_{2})$ is the norm limit of the truncated products, and for each truncation the finite double integral of $e^{λ_{1} z_{1} + λ_{2} z_{2}} R (λ_{1}) R (λ_{2})$ equals its iterate, so $T (z_{1}) T (z_{2}) = \frac{1}{(2 π i)^{2}} \int_{Γ_{2}} \int_{Γ_{1}} e^{λ_{1} z_{1} + λ_{2} z_{2}} R (λ_{1}) R (λ_{2}) d λ_{1} d λ_{2}$ with absolutely convergent iterated integrals.

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The finite-keyhole winding calculation. For $R > r$ , let $C_{R} (r, θ)$ be the closed contour obtained by appending to $Γ_{R} (r, θ)$ the counterclockwise outer arc $R e^{i α}$ , $θ \leq α \leq 2 π - θ$ , through the left half-plane. Its interior is $D_{R} = {λ : ∣ λ ∣ < R, ∣ λ ∣ < r or ∣ \arg λ ∣ > θ}$ , with $0$ included by the first alternative; it is star-shaped about $0$ . For $p$ off $C_{R}$ , the winding number is one when $p \in D_{R}$ and zero when $p$ is outside $\overline{D_{R}}$ ; equivalently $\oint_{C_{R} (r, θ)} (λ - p)^{- 1} d λ$ is $2 π i$ or $0$ , respectively. At $p = 0$ , the two radial integrals of $d λ / λ$ cancel, while the inner and outer arcs contribute $2 i θ$ and $i (2 π - 2 θ)$ , so the index is $1$ . Since $D_{R}$ is connected, [L5] gives the same index at every interior point. Every exterior point can move radially out beyond radius $R$ without meeting the trace, then along an outer circle, so it belongs to the unbounded component, where [L5] gives index $0$ . If $z \in Σ_{δ}$ and $θ$ is admissible for $z$ , then on the outer arc $Re (λ z) \leq - c R ∣ z ∣$ for some $c > 0$ , so $\int_{A_{R}} e^{λ z} (λ - p)^{- 1} d λ \to 0$ for fixed $p$ . Put $D_{\infty} : = ⋃_{R > r} D_{R}$ . If $p \in D_{\infty}$ , then $p \in D_{R}$ for every sufficiently large $R$ ; writing $e^{λ z} (λ - p)^{- 1} = e^{p z} (λ - p)^{- 1} + G (λ)$ with entire $G$ having a global primitive shows that the open contour integral tends to $2 π i e^{p z}$ . If instead $p \notin \overline{D_{\infty}}$ , then $p$ lies outside every $\overline{D_{R}}$ and, for each finite $R$ , choose $0 < ε_{R} < dist (p, \overline{D_{R}})$ and set $U_{R, ε_{R}} = \overline{D_{R}} + B (0, ε_{R})$ . This is an open star-shaped neighborhood of $\overline{D_{R}}$ that avoids $p$ ; the integrand is holomorphic there, so its closed integral vanishes by [L3], and the outer arc tends to zero, giving an open contour integral equal to zero. The nested-contour cases evaluated below have $p$ either inside $D_{\infty}$ or outside $\overline{D_{\infty}}$ , so no boundary-pole case is needed.

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The resolvent identity and the inner integrals. By [L2], $R (λ_{1}) R (λ_{2}) = \frac{R (λ_{2}) - R (λ_{1})}{λ_{1} - λ_{2}}$ , so the double integral of [step 1.1] splits as $A_{2} - A_{1}$ , where $A_{1}$ is the term with $R (λ_{1})$ and the inner $Γ_{2}$ integral, and $A_{2}$ is the term with $R (λ_{2})$ and the inner $Γ_{1}$ integral. The two infinite contours are disjoint and have positive separation $d : = dist (Γ_{1}, Γ_{2}) > 0$ : their finite arc pieces are disjoint compact sets, and their ray tails have distinct angles $θ_{1} \neq θ_{2}$ , so the distance between tails tends to infinity. Thus $∣ λ_{1} - λ_{2} ∣^{- 1} \leq d^{- 1}$ ; together with exponential decay along the rays and the sectorial resolvent bound, this gives absolute integrability of each split double-integral term and justifies Fubini. For each fixed $λ_{1} \in Γ_{1}$ , the nesting in [step 1.1] puts $λ_{1}$ outside the closure of the full unbounded keyhole region $D_{\infty, 2} : = ⋃_{R_{2} > r_{2}} D_{R_{2}} (r_{2}, θ_{2})$ , so it is outside every truncated interior. For each $R_{2}$ , the scalar integrand $e^{λ_{2} z_{2}} / (λ_{1} - λ_{2})$ is holomorphic on an open star-shaped neighborhood of $\overline{D_{R_{2}}}$ avoiding its pole, so the integral on $C_{R_{2}} (r_{2}, θ_{2})$ is zero by [L3]. Its outer arc contribution tends to zero by exponential decay, hence the open inner $Γ_{2}$ integral in $A_{1}$ vanishes. For each fixed $λ_{2} \in Γ_{2}$ , the nesting puts it inside the unbounded keyhole interior of $Γ_{1}$ , and therefore in $D_{R_{1}}$ for every sufficiently large $R_{1}$ . On $C_{R_{1}} (r_{1}, θ_{1})$ , the decomposition $e^{λ_{1} z_{1}} / (λ_{1} - λ_{2}) = e^{λ_{2} z_{1}} / (λ_{1} - λ_{2}) + [e^{λ_{1} z_{1}} - e^{λ_{2} z_{1}}] / (λ_{1} - λ_{2})$ has an entire second term with a global primitive; by [step 1.2] the first term integrates to $2 π i e^{λ_{2} z_{1}}$ . The outer arc of the original exponential integrand tends to zero for this fixed $λ_{2}$ , so the open inner $Γ_{1}$ integral in $A_{2}$ is $2 π i e^{λ_{2} z_{1}}$ . These are pointwise evaluations of the inner improper integrals; no radius-limit is interchanged with the outer integration.

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Strong continuity with the quantitative estimate. Let $x \in D (A)$ and fix $0 < δ^{'} < δ$ . Choose once and for all an angle $θ$ with $π / 2 + δ^{'} < θ < π / 2 + δ$ , and choose $ε > 0$ so that $θ < π / 2 + δ - ε$ ; this same contour angle is admissible for every $z \in Σ_{δ^{'}}$ . Every $μ$ on the contour is nonzero, so [L2] gives $R (μ, A) x - μ^{- 1} x = μ^{- 1} R (μ, A) A x$ , and the scalar identity $\frac{1}{2 π i} \int_{Γ} e^{μ z} \frac{d μ}{μ} = 1$ for $z \in Σ_{δ^{'}}$ follows by closing the truncated keyhole: the closed integral is $2 π i$ by [step 1.2] at $p = 0$ plus the entire quotient $(e^{μ z} - 1) / μ$ , whose integral is zero, and the outer arc of $e^{μ z} / μ$ tends to zero. Hence $T (z) x - x = \frac{1}{2 π i} \int_{Γ} e^{μ z} μ^{- 1} R (μ, A) A x d μ$ . Use this fixed $θ$ and the contour inner radius $ϱ = 1 / ∣ z ∣$ . Since $∣ \arg z ∣ < δ^{'}$ , on both rays $Re (μ z) \leq - c s ∣ z ∣$ with a single $c = c (θ, δ^{'}) > 0$ ; the two rays contribute at most $2 M_{ε} ∥ A x ∥ ∣ z ∣ \int_{1}^{\infty} e^{- c σ} σ^{- 2} d σ$ after $σ = s ∣ z ∣$ , and the arc contributes at most $2 θ e M_{ε} ∣ z ∣ ∥ A x ∥$ . Thus $∥ T (z) x - x ∥ \leq C_{δ^{'}, r} ∣ z ∣ ∥ A x ∥$ with one constant for all $z \in Σ_{δ^{'}}$ and $∣ z ∣ \leq r$ , and this tends to $0$ as $z \to 0$ in that smaller sector.

3.1step 1.1step 2.1L1givenalgebra

The semigroup law. Substituting [step 2.1] into [step 1.1] gives $T (z_{1}) T (z_{2}) = \frac{1}{(2 π i)^{2}} \int_{Γ_{2}} e^{λ_{2} z_{2}} 2 π i e^{λ_{2} z_{1}} R (λ_{2}, A) d λ_{2} = \frac{1}{2 π i} \int_{Γ_{2}} e^{λ_{2} (z_{1} + z_{2})} R (λ_{2}, A) d λ_{2} = T (z_{1} + z_{2})$ , the last equality by the independence of the contour [L1], since $Γ_{2}$ is admissible for $z_{1} + z_{2} \in Σ_{δ_{0}}$ when $θ_{2} > π / 2 + δ_{0}$ .

4.1step 2.2step 3.1L1given∎

Extension to all $x$ and assembly. $D (A)$ is dense in $X$ , and $∥ T (z) ∥ \leq M_{δ^{'}}$ on $Σ_{δ^{'}}$ by [L1], so the uniform estimate of [step 2.2] on the dense set extends the strong limit $T (z) x \to x$ to every $x \in X$ . Together with the semigroup law [step 3.1], the holomorphy and boundedness of [L1], and $T (0) = I$ , this exhibits $T$ as a bounded analytic semigroup of angle $δ$ ; the argument used only the resolvent identity, the contour computations and norm estimates, so no choice principle beyond Dependent Choice was used.

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Cauchy estimates for an analytic semigroup give generator power bounds

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 (Banach space) and let $(T (z))_{z \in Σ_{δ} \cup {0}}$ be a bounded analytic semigroup of angle $δ \in (0, π / 2]$ with generator $A$ (Complex sector and bounded analytic semigroup, Infinitesimal generator of a C0-semigroup), and put $M_{δ^{'}} : = \sup_{z \in Σ_{δ^{'}}} ∥ T (z) ∥ < \infty$ for $0 < δ^{'} < δ$ . Then for every $t > 0$ , every $m \geq 1$ and every $0 < δ^{''} < δ^{'} < δ$ :

$T (t) X \subseteq D (A^{m})$ and the identity $T^{(m)} (t) = A^{m} T (t)$ holds as bounded operators, where $T^{(m)}$ is the $m$ -th norm derivative on $(0, \infty)$ ;
$∥ A^{m} T (t) ∥ = ∥ T^{(m)} (t) ∥ \leq \frac{m! M_{δ^{'}}}{(t \sin δ^{''})^{m}} .$

No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A bounded analytic semigroup $T$ of angle $δ$ on a complex Banach space $X$ with generator $A$ , constants $M_{δ^{'}} < \infty$ for $δ^{'} < δ$ , times $t > 0$ , integers $m \geq 1$ , and angles $0 < δ^{''} < δ^{'} < δ$ .

[L1]

The family $T : Σ_{δ} \to B (X)$ is norm-holomorphic, $T (0) = I$ , $T (z_{1} + z_{2}) = T (z_{1}) T (z_{2})$ for $z_{1}, z_{2} \in Σ_{δ}$ , $∥ T (z) ∥ \leq M_{δ^{'}}$ for $z \in Σ_{δ^{'}}$ , and $\lim_{Σ_{δ^{'}} ∋ z \to 0} T (z) x = x$ for every $x \in X$ (Complex sector and bounded analytic semigroup).

[L2]

A vector $y \in X$ lies in $D (A)$ exactly when the strong right derivative $\lim_{h ↓ 0} h^{- 1} (T (h) y - y)$ exists, and then $A y$ is that limit; hence for $h > 0$ small and $y \in X$ one may test membership of $T (t) y$ in $D (A)$ by this limit (Infinitesimal generator of a C0-semigroup).

[L3]

For a continuous complex-differentiable $F : U \to Y$ into a complex Banach space $Y$ whose closed disc $\overline{D (z_{0}, R)}$ lies in $U$ , and every $0 < r < R$ , the $m$ -th derivative satisfies $∥ F^{(m)} (z_{0}) ∥ \leq m! \sup_{∣ w - z_{0} ∣ = r} ∥ F (w) ∥ / r^{m}$ , and all derivatives exist (Cauchy integral formula and Cauchy estimates for Banach-valued holomorphic functions).

[L4]

Since $X$ is Banach, $B (X)$ is Banach in the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach); thus [L3] applies to the $B (X)$ -valued map $T$ .

Proof

technique · direct

1.1L1L2givenalgebra

The first-order identity. Fix $t > 0$ and $y \in X$ . For $h > 0$ small, the semigroup law [L1] with $t, t + h \in Σ_{δ}$ gives $h^{- 1} (T (h) - I) T (t) y = h^{- 1} (T (t + h) y - T (t) y)$ , and the right-hand side converges as $h ↓ 0$ to the complex derivative $T^{'} (t) y$ of the holomorphic map $z \mapsto T (z) y$ at $t$ , because that derivative exists in operator norm by [L1]; hence by [L2] $T (t) y \in D (A)$ and $A T (t) y = T^{'} (t) y$ .

2.1step 1.1L1L2L3L4givenalgebra

The inductive identity. Assume $T (s) X \subseteq D (A^{m})$ and $A^{m} T (s) = T^{(m)} (s) \in B (X)$ for all $s > 0$ . Fix $t > 0$ and $y \in X$ , and put $x : = A^{m} T (t) y = T^{(m)} (t) y$ . For $h > 0$ , the semigroup law [L1] gives $T (h) T (t + s) = T (t + h + s)$ for real $s$ near $0$ ; differentiating this identity $m$ times in operator norm, justified by [L3, L4], gives $T (h) T^{(m)} (t) y = T^{(m)} (t + h) y$ . Therefore $\frac{T (h) x - x}{h} = \frac{T^{(m)} (t + h) y - T^{(m)} (t) y}{h} ⟶ T^{(m + 1)} (t) y (h ↓ 0) .$ By the generator definition [L2], $x \in D (A)$ and $A x = T^{(m + 1)} (t) y$ . Since $T (t) y \in D (A^{m})$ by the induction hypothesis and $A^{m} T (t) y = x \in D (A)$ , the recursive definition of powers gives $T (t) y \in D (A^{m + 1})$ and $A^{m + 1} T (t) y = T^{(m + 1)} (t) y$ . As $y$ was arbitrary, $T (t) X \subseteq D (A^{m + 1})$ and $A^{m + 1} T (t) = T^{(m + 1)} (t) \in B (X)$ .

3.1step 2.1given

Conclusion of the identity. [step 1.1] is the case $m = 1$ and [step 2.1] carries every higher $m$ , so $T (t) X \subseteq D (A^{m})$ and $T^{(m)} (t) = A^{m} T (t) \in B (X)$ for every $t > 0$ and $m \geq 1$ .

4.1step 3.1L1L3L4givenalgebra∎

The Cauchy estimate. Fix $t > 0$ and $0 < δ^{''} < δ^{'} < δ$ . Choose $r : = t \sin δ^{''}$ and $R$ with $r < R < t \sin δ^{'}$ . The closed disc $\overline{D (t, R)}$ lies in $Σ_{δ^{'}}$ : its radius is smaller than the distance $t \sin δ^{'}$ from $t$ to either boundary ray, and $R < t$ keeps it away from the vertex. In particular the circle $∣ w - t ∣ = r$ lies in $Σ_{δ^{'}}$ , where [L1] bounds $∥ T (w) ∥$ by $M_{δ^{'}}$ . Applying the Cauchy estimate [L3] to the $B (X)$ -valued holomorphic map $F = T$ at $z_{0} = t$ gives $∥ T^{(m)} (t) ∥ \leq m! \sup_{∣ w - t ∣ = r} ∥ T (w) ∥ / r^{m} \leq m! M_{δ^{'}} / (t \sin δ^{''})^{m}$ , and with [step 3.1] this is $∥ A^{m} T (t) ∥ \leq m! M_{δ^{'}} / (t \sin δ^{''})^{m}$ .

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Smoothing estimates for the semigroup generated by a sectorial 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 $A$ be sectorial of angle $δ \in (0, π / 2]$ with vertex $0$ on a complex Banach space $X$ (Sectorial operator with the semigroup sign convention) and let $(T (t))_{t \geq 0}$ be the contour semigroup of The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex. Then for every $t > 0$ and every $m \geq 1$ :

$T (t) X \subseteq D (A^{m})$ and $∥ A^{m} T (t) ∥ \leq C_{m} t^{- m}$ , where $C_{m}$ depends only on $m$ , $δ$ and the sectoriality constants $M_{ε}$ ;
$T \in C^{\infty} ((0, \infty), B (X))$ in the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) and $\frac{d^{m}}{d t^{m}} T (t) = A^{m} T (t)$ for every $m \geq 1$ .

No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with vertex $0$ on a complex Banach space $X$ and its contour semigroup $T$ , with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ on $∣ \arg λ ∣ < π / 2 + δ - ε$ ; a contour $Γ = Γ (r, θ)$ with $r > 0$ , $θ = π / 2 + δ / 2$ , and $ε = δ / 4$ (so $θ < π / 2 + δ - ε$ ); a fixed $t > 0$ and $m \geq 1$ .

[L1]

For a fixed $Γ (r, θ)$ , the representation $T (z) = \frac{1}{2 π i} \int_{Γ} e^{λ z} R (λ, A) d λ$ converges absolutely and locally uniformly on compact subsets of $Σ_{θ - π / 2}$ . For general $z \in Σ_{δ}$ the angle must satisfy $θ > π / 2 + ∣ \arg z ∣$ . The resulting $T$ is norm-holomorphic on $Σ_{δ}$ and independent of the radius and admissible angle (The Dunford contour integral defines a bounded holomorphic family on the sector, The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex); convergence of the polynomial-weighted integrals at positive real times is proved in step 1.1 below.

[L2]

$A$ is closed, and for $λ \in ρ (A)$ one has $R (λ, A) X = D (A)$ and $A R (λ, A) = λ R (λ, A) - I$ on $X$ (Sectorial operator with the semigroup sign convention, Resolvent and spectrum of a closed operator on a Banach space).

[L3]

For the curve integral, $∥ \int_{γ} f ∥ \leq \int_{γ} ∥ f ∥$ and the integral is linear (Bochner integral norm inequality, Linearity of the Bochner integral).

Proof

technique · direct

1.1L1L3givenalgebra

The differentiated contour formula. For $t > 0$ and every $k \geq 0$ the integral $\frac{1}{2 π i} \int_{Γ} λ^{k} e^{λ t} R (λ, A) d λ$ converges absolutely, because on the rays $∥ λ^{k} e^{λ t} R (λ, A) ∥ \leq M_{ε} s^{k - 1} e^{- c s t}$ with $c > 0$ (the admissible angle gives $Re (λ t) \leq - c s t$ ) and $\int_{r}^{\infty} s^{k - 1} e^{- c s t} d s < \infty$ for $t > 0$ ; differentiating $k$ times under the integral sign is justified by the same integrable majorant on compact time intervals bounded away from $0$ , so $\frac{d^{k}}{d t^{k}} T (t) = \frac{1}{2 π i} \int_{Γ} λ^{k} e^{λ t} R (λ, A) d λ$ and $T \in C^{\infty} ((0, \infty), B (X))$ in operator norm.

1.2L1L2L3givenalgebra

The truncated integral lies in the domain. For $k \geq 0$ the truncated integral $I_{k, R} : = \frac{1}{2 π i} \int_{Γ_{R}} λ^{k} e^{λ t} R (λ, A) d λ$ is the norm limit of Riemann sums of elements of $D (A)$ , and by [L2] $A$ of each such sum equals the corresponding sum of $λ^{k + 1} e^{λ t} R (λ, A) - λ^{k} e^{λ t} I$ ; since $A$ is closed, the limit $I_{k, R}$ lies in $D (A)$ with $A I_{k, R} = \frac{1}{2 π i} \int_{Γ_{R}} λ^{k + 1} e^{λ t} R (λ, A) d λ - \frac{1}{2 π i} (\int_{Γ_{R}} λ^{k} e^{λ t} d λ) I$ .

1.3L1givenalgebra

The scalar contour integrals vanish. For every $k \geq 0$ one has $\int_{Γ} λ^{k} e^{λ t} d λ = 0$ : close the truncated contour by the arc at radius $R$ through the left, on which $Re (λ t) \leq - c R t$ ; the integral of the entire function $λ^{k} e^{λ t}$ over the closed truncated curve vanishes (it has the global primitive $\int_{0}^{λ} w^{k} e^{w t} d w$ ), while the closing arc contribution is at most $2 π R^{k + 1} e^{- c R t} \to 0$ .

2.1step 1.1step 1.2step 1.3L2givenalgebra

First order: $A T (t) = T^{'} (t)$ . Apply [step 1.2] with $k = 0$ : the truncated integral $I_{0, R}$ converges in operator norm to $T (t)$ , while its $A$ -image is $I_{1, R} - \frac{1}{2 π i} (\int_{Γ_{R}} e^{λ t} d λ) I$ . The first term converges in operator norm to $T^{'} (t)$ by [step 1.1], and the scalar term tends to zero by [step 1.3]. Since $A$ is closed by [L2], for each $x \in X$ the convergence of $I_{0, R} x$ and $A I_{0, R} x$ gives $T (t) x \in D (A)$ and $A T (t) x = T^{'} (t) x$ . Thus $T (t) X \subseteq D (A)$ and $A T (t) = T^{'} (t)$ .

3.1step 2.1step 1.1step 1.2step 1.3L2givenalgebra

Higher orders by the same argument. Suppose $T (t) X \subseteq D (A^{m})$ and $A^{m} T (t) = T^{(m)} (t)$ . Apply [step 1.2] with $k = m$ to the truncated contour integral $I_{m, R}$ for $T^{(m)}$ : its $A$ -image is $I_{m + 1, R}$ minus the scalar term in [step 1.2]. As $R \to \infty$ , $I_{m, R} \to T^{(m)} (t) = A^{m} T (t)$ and $I_{m + 1, R} \to T^{(m + 1)} (t)$ in operator norm by [step 1.1], while the scalar term tends to zero by [step 1.3]. Closedness of $A$ then gives $A^{m + 1} T (t) x = T^{(m + 1)} (t) x$ for every $x \in X$ , and $T (t) X \subseteq D (A^{m + 1})$ . Induction proves the claim for every $m \geq 1$ .

4.1step 1.1step 3.1L1L3givenalgebra∎

The bound. By the independence of the inner radius [L1], compute $T^{(m)} (t)$ with $r = 1 / t$ . On the rays, $∥ λ^{m} e^{λ t} R (λ, A) ∥ \leq M_{ε} s^{m - 1} e^{- c s t}$ and $\int_{1 / t}^{\infty} s^{m - 1} e^{- c s t} d s = t^{- m} \int_{1}^{\infty} σ^{m - 1} e^{- c σ} d σ$ after $σ = s t$ ; on the arc $∣ λ ∣ = 1 / t$ one has $∣ λ ∣^{m} = t^{- m}$ , length at most $2 θ / t$ , $∣ e^{λ t} ∣ \leq e$ and $∥ R (λ, A) ∥ \leq M_{ε} t$ , so the arc contributes at most $2 θ e M_{ε} t^{- m}$ . Hence $∥ A^{m} T (t) ∥ = ∥ T^{(m)} (t) ∥ \leq C_{m} t^{- m}$ with $C_{m}$ depending only on $m, δ, M_{δ / 4}$ through the prescribed $θ = π / 2 + δ / 2$ and $c = - \cos θ$ , and [step 3.1] supplies the domain membership and the derivative identity for every $m$ .

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Analytic semigroups are operator-norm differentiable away from zero

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.

In the setting of Smoothing estimates for the semigroup generated by a sectorial operator, the map $t \mapsto T (t) \in B (X)$ (A bounded linear operator between normed spaces) is of class $C^{\infty}$ on $(0, \infty)$ in the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum), with $\frac{d}{d t} T (t) = A T (t)$ for every $t > 0$ . In particular $T$ is immediately operator-norm differentiable on $(0, \infty)$ ; no norm continuity or differentiability at $0$ is asserted, and for an unbounded generator $A$ it fails (the heat counterexample of the companion page). No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with vertex $0$ on a complex Banach space $X$ and its contour semigroup $T$ , together with the conclusions of the smoothing theorem.

[L1]

$T \in C^{\infty} ((0, \infty), B (X))$ in the operator norm and $\frac{d^{m}}{d t^{m}} T (t) = A^{m} T (t)$ for every $t > 0$ and $m \geq 1$ (Smoothing estimates for the semigroup generated by a sectorial operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L2]

For any strongly continuous semigroup with generator $G$ , $J_{h} y : = \int_{0}^{h} T (s) y d s$ belongs to $D (G)$ and $G J_{h} y = T (h) y - y$ (Time integrals of semigroup orbits lie in the generator domain). A bounded operator within norm distance $1$ of $I$ is invertible by the Neumann series (Neumann series and small perturbations of bounded inverses).

Proof

technique · direct

1.1L1given

$C^{\infty}$ regularity and the first derivative. By [L1] the map $t \mapsto T (t)$ has norm derivatives of every order on $(0, \infty)$ and $\frac{d^{m}}{d t^{m}} T (t) = A^{m} T (t)$ for every $m \geq 1$ ; taking $m = 1$ gives $\frac{d}{d t} T (t) = A T (t)$ as bounded operators, and taking all $m$ gives the $C^{\infty}$ statement in the operator norm.

2.1step 1.1L1L2givenalgebra∎

The vertex and bounded generators. If $∥ T (t) - I ∥ \to 0$ as $t ↓ 0$ , choose $h > 0$ with $\sup_{0 \leq s \leq h} ∥ T (s) - I ∥ < 1 / 2$ . By [L2], the bounded operator $K_{h} : = J_{h} / h$ satisfies $∥ K_{h} - I ∥ \leq \sup_{0 \leq s \leq h} ∥ T (s) - I ∥ < 1 / 2$ and is invertible. Since its range lies in $D (G)$ , this forces $D (G) = X$ , and $G = (T (h) - I) K_{h}^{- 1} / h$ is bounded. Thus an unbounded generator cannot have norm continuity at the vertex. Step 1.1 gives the asserted positive-time regularity, and this argument proves the general exclusion at zero rather than inferring it from one heat example.

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Sectorial resolvent characterisation of bounded analytic semigroups

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 $A$ be a closed densely defined linear operator on $X$ (Densely defined, closed and closable operators, and cores, Resolvent and spectrum of a closed operator on a Banach space). The following are equivalent:

(a) $A$ has a bounded analytic semigroup extension to some sector $Σ_{δ}$ , $δ > 0$ (Complex sector and bounded analytic semigroup);

(b) there is $ϑ \in (0, π / 2)$ such that both $e^{i ϑ} A$ and $e^{- i ϑ} A$ , on the common domain $D (A)$ , generate bounded strongly continuous semigroups;

(c) $A$ generates a bounded strongly continuous semigroup $(T (t))_{t \geq 0}$ with $T (t) X \subseteq D (A)$ for all $t > 0$ and $\sup_{t > 0} ∥ t A T (t) ∥ < \infty$ ;

(d) $A$ generates a bounded strongly continuous semigroup and there is $C > 0$ such that $∥ R (r + i s, A) ∥ \leq C / ∣ s ∣$ for every $r > 0$ and $s \neq 0$ ;

(e) $A$ satisfies the sectorial resolvent condition with vertex $0$ for some positive exponent in the sense of Sectorial operator with the semigroup sign convention.

If these conditions hold, the semigroup in (c) is the contour semigroup. The maximal analytic angle equals the supremum of admissible rotation angles in (b) and the supremum of sectorial exponents in (e); these are supremal exponents, not arbitrary smaller witnesses. No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A closed densely defined linear operator $A$ on the complex Banach space $X$ ; the definitions of a strongly continuous semigroup and its generator, of a bounded analytic semigroup, and of sectoriality at vertex $0$ ; and, whenever one of (a)-(d) is assumed below, the corresponding semigroup with its constants.

[L1]

A bounded analytic semigroup on $Σ_{δ} \cup {0}$ is a family with $T (0) = I$ , $T (z_{1} + z_{2}) = T (z_{1}) T (z_{2})$ , operator-norm holomorphy on $Σ_{δ}$ , strong continuity at the vertex, and uniform boundedness on every strictly smaller sector; its generator is the infinitesimal generator of the strongly continuous semigroup $(T (t))_{t \geq 0}$ (Complex sector and bounded analytic semigroup).

[L2]

$A$ is sectorial of angle $δ \in (0, π / 2]$ at vertex $0$ if $Σ_{π / 2 + δ} \subseteq ρ (A)$ and for every $ε \in (0, δ)$ there is $M_{ε} \geq 1$ with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ on $Σ_{π / 2 + δ - ε}$ ; $ρ (A)$ is the set of $λ$ for which $λ I - A$ is bijective and $R (λ, A) = (λ I - A)^{- 1}$ , with $R (λ, A) X = D (A)$ and $A R (λ, A) = λ R (λ, A) - I$ (Sectorial operator with the semigroup sign convention, Resolvent and spectrum of a closed operator on a Banach space).

[L3]

For a semigroup supplied with an exponential bound, closedness and density are established in step 1.1 below. For $y \in D (B)$ one has $B S (t) y = S (t) B y$ for all $t \geq 0$ (The generator commutes with the semigroup on its domain, Infinitesimal generator of a C0-semigroup).

[L4]

The fundamental theorem of calculus for Banach-valued continuous curves and average convergence: for continuous $g$ one has $\frac{1}{h} \int_{t}^{t + h} g \to g (t)$ as $h ↓ 0$ (Fundamental theorem of calculus for Banach-valued continuous curves, Average convergence for a continuous Banach-valued function, Bochner-integrable function).

[L5]

Banach-valued Cauchy theorem on a star-shaped open set $U$ : a continuous complex-differentiable $F : U \to Y$ has $\int_{γ} F d w = 0$ for every closed piecewise $C^{1}$ contour in $U$ (Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains).

[L6]

For a sectorial $A$ of angle $δ \in (0, π / 2]$ the contour family $T (z) = \frac{1}{2 π i} \int_{Γ} e^{λ z} R (λ, A) d λ$ is a bounded analytic semigroup of angle $δ$ with generator $A$ , unique among exponentially bounded semigroups with that generator, and it satisfies $T (t) X \subseteq D (A^{m})$ and $∥ A^{m} T (t) ∥ \leq C_{m} t^{- m}$ for all $m \geq 1$ , $t > 0$ (The Dunford contour integral defines a bounded holomorphic family on the sector, The Dunford contour construction satisfies the semigroup law and strong continuity at the vertex, The generator of the contour semigroup is the sectorial operator, Smoothing estimates for the semigroup generated by a sectorial operator).

[L7]

A holomorphic Banach-space-valued function on a disc has a norm-convergent power-series expansion there, and two power series about a real centre that agree on a real interval have equal coefficients and hence equal sums on the disc (Cauchy integral formula and Cauchy estimates for Banach-valued holomorphic functions, Banach-valued power series are determined by their real values).

[L8]

Taylor's formula with integral remainder holds for $C^{n + 1}$ curves into a Banach space on real intervals (Taylor expansion with integral remainder for Banach-valued curves).

[L9]

Resolvent identities: $R (λ, B) - R (μ, B) = (μ - λ) R (λ, B) R (μ, B)$ , and $R (\cdot, B)$ is norm-holomorphic on $ρ (B)$ (Resolvent identity and holomorphy for a closed operator).

Proof

technique · direct

1.1L3L4givenalgebra

Complex Laplace representation. Let $(S (s))_{s \geq 0}$ be a strongly continuous semigroup with generator $B$ and $∥ S (s) ∥ \leq M e^{ω s}$ , and fix $λ \in C$ with $Re λ > ω$ ; the integral $v : = \int_{0}^{\infty} e^{- λ s} S (s) x d s$ converges absolutely with $∥ v ∥ \leq M (Re λ - ω)^{- 1} ∥ x ∥$ , and $(S (h) - I) v / h = ((e^{λ h} - 1) / h) \int_{h}^{\infty} e^{- λ s} S (s) x d s - h^{- 1} \int_{0}^{h} e^{- λ s} S (s) x d s \to λ v - x$ by absolute convergence and average convergence at $0$ , so $v \in D (B)$ and $B v = λ v - x$ ; if $y \in D (B)$ has $B y = λ y$ then $s \mapsto e^{- λ s} S (s) y$ has derivative $e^{- λ s} S (s) (B y - λ y) = 0$ by the commutation lemma and the fundamental theorem, so $∥ y ∥ = e^{- Re λ s} ∥ S (s) y ∥ \leq M e^{(ω - Re λ) s} ∥ y ∥ \to 0$ ; and for $y \in D (B)$ the same computation gives $\int_{0}^{\infty} e^{- λ s} S (s) (λ y - B y) d s = y$ , so the bounded linear map $x \mapsto v$ is a two-sided inverse of $λ I - B$ and $λ \in ρ (B)$ with $R (λ, B) x = v$ . Its graph is closed by continuity; swapping coordinates shows that $λ I - B$ has closed graph, and the continuous coordinate change $(y, w) \mapsto (y, λ y - w)$ proves that $B$ is closed. The integrated-orbit identity of Time integrals of semigroup orbits lie in the generator domain puts $h^{- 1} \int_{0}^{h} S (s) x d s$ in $D (B)$ ; average convergence [L4] shows that these vectors tend to every $x$ , so $D (B)$ is dense.

1.2L2L9givenalgebra

Resolvent scaling and agreement. For a closed linear operator $B$ , $c \neq 0$ and $λ \in ρ (B)$ one has $c λ \in ρ (c B)$ with $R (c λ, c B) = c^{- 1} R (λ, B)$ , because $c λ I - c B = c (λ I - B)$ and $D (c B) = D (B)$ , and conversely; if $S, T$ are closed operators and $R (λ, S) = R (λ, T)$ for one $λ$ , then $(λ I - S)^{- 1} = (λ I - T)^{- 1}$ gives $λ I - S = λ I - T$ and $D (S) = R (λ, S) X = R (λ, T) X = D (T)$ , hence $S = T$ .

1.3L1givenalgebra

Rotations of an analytic semigroup. Assume (a), so $T$ is a bounded analytic semigroup on $Σ_{δ} \cup {0}$ with generator $A$ ; for $0 < ϑ < δ$ and $s \geq 0$ put $T_{ϑ} (s) : = T (e^{i ϑ} s)$ and $T_{- ϑ} (s) : = T (e^{- i ϑ} s)$ : then $T_{\pm ϑ} (0) = I$ , $T_{\pm ϑ} (s + s^{'}) = T_{\pm ϑ} (s) T_{\pm ϑ} (s^{'})$ because $e^{\pm i ϑ} (s + s^{'}) = e^{\pm i ϑ} s + e^{\pm i ϑ} s^{'}$ , each orbit is continuous on $[0, \infty)$ by holomorphy on the sector and strong continuity at the vertex, and $∥ T_{\pm ϑ} (s) ∥ \leq \sup_{z \in Σ_{δ^{'}}} ∥ T (z) ∥ < \infty$ for any $ϑ < δ^{'} < δ$ ; thus both are bounded strongly continuous semigroups.

1.4L3L4L8givenalgebra

The (c) calculus and the local series. Assume (c), and put $M_{0} : = \sup_{t \geq 0} ∥ T (t) ∥$ and $M_{1} : = \sup_{t > 0} t ∥ A T (t) ∥$ . For $B_{s} : = A T (s)$ , the domain inclusion in (c) and generator commutation [L3] give $B_{s} T (s) = A T (2 s) = T (s) B_{s}$ , hence $B_{s}$ commutes with every $T (n s)$ and $B_{(n + 1) s} = B_{s} T (n s)$ . Induction yields $T (t) X \subseteq D (A^{n})$ and $A^{n} T (t) = B_{t / n}^{n}$ , with $∥ A^{n} T (t) ∥ \leq (M_{1} n / t)^{n}$ . First, $T$ is locally Lipschitz in operator norm on $(0, \infty)$ . Fix $0 < a < b$ and $0 < h \leq a / 2$ . Since $T (s) X \subseteq D (A)$ , the orbit formula and the fundamental theorem [L3, L4] give, for $a \leq s \leq b$ , $T (s + h) x - T (s) x = \int_{0}^{h} T (r) A T (s) x d r,$ so $∥ T (s + h) - T (s) ∥ \leq h \sup_{0 \leq r \leq a / 2} ∥ T (r) ∥ M_{1} / a$ . The same estimate applied from $s + h$ to $s$ handles negative increments on compact subintervals of $(0, \infty)$ . Thus $T$ is locally norm-continuous. For $n \geq 1$ and $0 < s_{1} < s_{2}$ , the semigroup law gives $A^{n} T (s_{j}) = A^{n} T (s_{1} / 2) T (s_{j} - s_{1} / 2), j = 1, 2.$ The power bound just proved and norm continuity of $T$ show that $s \mapsto A^{n} T (s)$ is norm-continuous on every compact subinterval of $(0, \infty)$ ; this is also true for $n = 0$ . For each $n \geq 0$ and $h > 0$ , the generic generator-orbit formula and commutation with $A^{n}$ give $A^{n} T (s + h) x - A^{n} T (s) x = \int_{s}^{s + h} A^{n + 1} T (σ) x d σ .$ For $h < 0$ the same formula follows by reversing the endpoints. Since $A^{n + 1} T (σ)$ is operator-norm continuous, division by $h$ shows $\frac{d}{d s} (A^{n} T (s)) = A^{n + 1} T (s)$ in operator norm. Hence $T \in C^{\infty} ((0, \infty), B (X))$ and $T^{(n)} = A^{n} T$ . If $M_{1} > 0$ , set $ρ : = \min {1 / 2, 1 / (2 e M_{1})}$ . Taylor's formula on $[t - ∣ h ∣, t + ∣ h ∣]$ then gives $∥ T (t + h) - \sum_{n < N} h^{n} A^{n} T (t) / n! ∥ \leq (e M_{1} ∣ h ∣ / (t - ∣ h ∣))^{N} \to 0$ for $∣ h ∣ < ρ t$ . The zeroth series term has norm at most $M_{0}$ , and for $n \geq 1$ the power bound gives $∣ z - t ∣^{n} ∥ A^{n} T (t) ∥ / n! \leq (ρ e M_{1})^{n}$ ; hence $\sum_{n \geq 0} ∣ z - t ∣^{n} ∥ A^{n} T (t) ∥ / n! \leq M_{0} + \sum_{n \geq 1} (ρ e M_{1})^{n} \leq M_{0} + 1$ for $∣ z - t ∣ \leq ρ t$ . If $M_{1} = 0$ , then $A T (s) = 0$ for all $s > 0$ ; generator commutation and the fundamental theorem [L3, L4] give $T (t) x = x$ for every $x \in D (A)$ , and density gives $T (t) = I$ and $A = 0$ .

1.5L7givenalgebra

Banach-valued identity theorem. Let $F : U \to Y$ be holomorphic on a connected open set. If $F$ vanishes on a real interval, choose a real center and a disc whose real diameter lies in that interval. The power series in [L7] then has all coefficients zero, so $F$ vanishes on a disc. If $F$ vanishes on any nonempty open set, put $Z = {z \in U : F vanishes on a neighborhood of z}$ . This set is nonempty and open. At every point in its closure in $U$ , continuity of every derivative from [L7] gives $F^{(n)} (z) = 0$ for every $n$ , since all derivatives vanish on $Z$ . The Taylor expansion at $z$ therefore vanishes on a neighborhood, so $z \in Z$ . Thus $Z$ is also closed; connectedness gives $Z = U$ .

1.6L2L3L9givenalgebra

(d) gives a wedge at the imaginary axis. Assume (d) with constants $M, C$ ; for fixed $s \neq 0$ the resolvent identity gives $∥ R (r + i s, A) - R (r^{'} + i s, A) ∥ \leq ∣ r - r^{'} ∣ C^{2} / s^{2}$ for $r, r^{'} > 0$ , so $R (r + i s, A)$ converges in norm as $r ↓ 0$ to some $R_{0} \in B (X)$ ; the vectors $y_{r} : = R (r + i s, A) x$ satisfy $((r + i s) I - A) y_{r} = x$ , so $A y_{r} = (r + i s) y_{r} - x \to i s R_{0} x - x$ and closedness of $A$ gives $R_{0} x \in D (A)$ with $(i s I - A) R_{0} x = x$ ; if $(i s I - A) x = 0$ then $x = R (r + i s, A) (r x) = r R (r + i s, A) x$ and $∥ x ∥ \leq C r ∣ s ∣^{- 1} ∥ x ∥$ for all $r > 0$ , so $x = 0$ ; hence $i s \in ρ (A)$ and $∥ R (i s, A) ∥ \leq C / ∣ s ∣$ , and the Neumann series $R (λ, A) = [I + r R (i s, A)]^{- 1} R (i s, A)$ converges for $λ = r + i s$ with $∣ r ∣ < q ∣ s ∣ / C$ , $q \in (0, 1)$ , giving $∥ R (λ, A) ∥ \leq C / ((1 - q) \cos θ ∣ λ ∣)$ , $θ : = \arctan (q / C)$ , on that wedge.

1.7L6givenalgebra

(e) gives the contour semigroup. Assume (e), so $A$ is sectorial of some angle $δ_{0} > 0$ ; putting $δ : = \min {δ_{0}, π / 2}$ (a smaller sector is contained in a larger one, so $A$ is sectorial of angle $δ$ ) and applying [L6], the contour family is a bounded analytic semigroup of angle $δ$ with generator $A$ , unique among exponentially bounded semigroups with that generator, and satisfies $T (t) X \subseteq D (A^{m})$ , $∥ A^{m} T (t) ∥ \leq C_{m} t^{- m}$ for all $m \geq 1$ , $t > 0$ ; in particular this semigroup satisfies (c) and supplies the semigroup of (a).

2.1step 1.1step 1.2step 1.3L2L3L5givenalgebra

(a) implies (b). Assume (a) and fix $0 < ϑ < δ$ ; by [step 1.1] with $λ = 1$ one has $R (1, A) x = \int_{0}^{\infty} e^{- t} T (t) x d t$ for every $x$ , and the function $z \mapsto e^{- z} T (z) x$ is holomorphic on the star-shaped sector $Σ_{δ}$ , so [L5] applied to the closed contours $ε \to R \to R e^{i ϑ} \to ε e^{i ϑ} \to ε$ gives $0 = \int_{ε}^{R} e^{- t} T (t) x d t + \int_{{a r c}_{R}} + \int_{r a y} + \int_{{a r c}_{ε}}$ with $∥ e^{- z} T (z) x ∥ \leq e^{- Re z} M_{δ^{'}} ∥ x ∥$ ; since $Re (s e^{i ϑ}) \geq s \cos ϑ$ and $Re z \geq R \cos ϑ$ on the outer arc and $∥ e^{- z} ∥ \leq e^{ε}$ on the inner arc, the limits $ε ↓ 0$ , $R \to \infty$ give $\int_{0}^{\infty} e^{- t} T (t) x d t = e^{i ϑ} \int_{0}^{\infty} e^{- s e^{i ϑ}} T_{ϑ} (s) x d s$ ; by [step 1.1] applied to the bounded semigroup $T_{ϑ}$ at $λ = e^{i ϑ}$ the last integral equals $R (e^{i ϑ}, A_{ϑ}) x$ , so $R (1, A) = e^{i ϑ} R (e^{i ϑ}, A_{ϑ}) = R (1, e^{- i ϑ} A_{ϑ})$ by [step 1.2]; both $A$ and $e^{- i ϑ} A_{ϑ}$ are closed by the hypothesis and step 1.1, so [step 1.2] gives $A = e^{- i ϑ} A_{ϑ}$ , that is $A_{ϑ} = e^{i ϑ} A$ with domain $D (A)$ ; the same argument with $- ϑ$ gives $A_{- ϑ} = e^{- i ϑ} A$ , so (b) holds for this $ϑ$ and hence for every $ϑ \in (0, δ)$ .

2.2step 1.1step 1.2L2givenalgebra

(b) implies (e). Assume (b) for some $ϑ$ , with the two bounded semigroups satisfying $∥ S_{\pm} (t) ∥ \leq M$ ; [step 1.1] applied to $S_{\pm}$ gives $C_{+} \subseteq ρ (e^{\pm i ϑ} A)$ with $∥ R (λ, e^{\pm i ϑ} A) ∥ \leq M / Re λ$ , so by [step 1.2] the half-planes ${λ : Re (e^{\pm i ϑ} λ) > 0} = e^{\mp i ϑ} C_{+}$ lie in $ρ (A)$ ; their union is $Σ_{π / 2 + ϑ}$ , and for $λ = r e^{i α}$ with $∣ α ∣ \leq π / 2 + ϑ - ε$ the sign $α \geq 0$ gives $Re (e^{- i ϑ} λ) = r \cos (α - ϑ) \geq r \min {\cos ϑ, \sin ε}$ (the angle $α - ϑ$ ranges over $[- ϑ, π / 2 - ε]$ ) and symmetrically for $α < 0$ , so $∥ R (λ, A) ∥ \leq M / (r \min {\cos ϑ, \sin ε}) = M_{ε} / ∣ λ ∣$ there; hence $A$ is sectorial of angle $ϑ$ , which is (e).

2.3step 1.1step 1.6L2givenalgebra

(d) implies (e). Assume (d); [step 1.6] gives the wedge $∣ r ∣ < q ∣ s ∣ / C$ with $∥ R (λ, A) ∥ \leq C / ((1 - q) \cos θ ∣ λ ∣)$ , and [step 1.1] applied to the bounded semigroup generated by $A$ gives $∥ R (λ, A) ∥ \leq M / Re λ$ on the right half-plane; set $η : = \frac{1}{2} \arctan (q / C) > 0$ : for $∣ \arg λ ∣ \leq π / 2 + η - ε$ either $∣ \arg λ ∣ \leq π / 2 - ε / 2$ and the Laplace estimate gives $∥ R (λ, A) ∥ \leq M / (∣ λ ∣ \sin (ε / 2))$ , or $∣ \arg λ ∣ > π / 2 - ε / 2$ and then $∣ r ∣ / ∣ s ∣ \leq \tan η < q / C$ , so the wedge bound applies; hence $ρ (A) \supseteq Σ_{π / 2 + η}$ and the bound $M_{ε} / ∣ λ ∣$ holds on each $Σ_{π / 2 + η - ε}$ , which is (e) with exponent $η$ .

2.4step 1.4step 1.5L7givenalgebra

(c) builds the holomorphic extension. Assume (c) and $M_{1} > 0$ , put $ρ : = \min {1 / 2, 1 / (2 e M_{1})}$ and $Ω : = ⋃_{t > 0} D (t, ρ t)$ ; by [step 1.4] each series $S_{t} (z) : = \sum_{n \geq 0} (z - t)^{n} A^{n} T (t) / n!$ converges in $B (X)$ for $∣ z - t ∣ < t / (e M_{1})$ , is bounded by $M_{0} + 1$ on $D (t, ρ t)$ , and equals $T$ on the real interval $(t - ρ t, t + ρ t)$ ; if $z \in D (t_{1}, ρ t_{1}) \cap D (t_{2}, ρ t_{2})$ the two series are holomorphic on the convex intersection and agree on its nonempty real interval, hence agree on it by [step 1.5], so $\tilde{T} (z) : = S_{t} (z)$ is a well-defined holomorphic map $Ω \to B (X)$ with $∥ \tilde{T} ∥ \leq M_{0} + 1$ extending $T$ ; for real $t > 0$ the holomorphic difference $z \mapsto \tilde{T} (z) T (t) - \tilde{T} (z + t)$ vanishes on a real interval and hence, by [step 1.5], on every connected component of ${z : z, z + t \in Ω}$ meeting it, so $\tilde{T} (z) T (t) = \tilde{T} (z + t)$ for $z$ in the sector $Σ_{η}$ , $η : = \arctan ρ > 0$ , which lies in $Ω$ because $∣ Im z ∣ < ρ Re z$ ; if $M_{1} = 0$ then $T (t) = I$ and $A = 0$ by [step 1.4], and $\tilde{T} (z) : = I$ is a bounded analytic semigroup of angle $π / 2$ with generator $A$ .

3.1step 1.5step 2.4L1givenalgebra

(c) implies (a). Restrict the holomorphic extension of step 2.4 to the connected sector $Σ_{η}$ , where $\tilde{T}$ is bounded by $M_{0} + 1$ . For each fixed real $τ > 0$ , the maps $z \mapsto T (τ) \tilde{T} (z)$ and $z \mapsto \tilde{T} (z) T (τ)$ are holomorphic on $Σ_{η}$ . They agree for every positive real $s$ , because $\tilde{T} (s) = T (s)$ and the real semigroup law gives $T (τ) T (s) = T (τ + s) = T (s + τ) = T (s) T (τ)$ . By the Banach-valued identity theorem [step 1.5], they agree throughout $Σ_{η}$ . Combining this commutation with step 2.4 yields, for every $z \in Σ_{η}$ and real $τ > 0$ , $T (τ) \tilde{T} (z) = \tilde{T} (z) T (τ) = \tilde{T} (z + τ)$ . Now fix $w \in Σ_{η}$ . Since the sector is closed under addition, $G (z) : = \tilde{T} (z) \tilde{T} (w) - \tilde{T} (z + w)$ is holomorphic on $Σ_{η}$ ; for real $z = t > 0$ the mixed-product identity just proved gives $G (t) = T (t) \tilde{T} (w) - \tilde{T} (t + w) = 0$ . A second application of [step 1.5] gives $G \equiv 0$ , so $\tilde{T} (z) \tilde{T} (w) = \tilde{T} (z + w)$ . Set $\tilde{T} (0) : = I$ . To check strong continuity at the vertex, for fixed small real $h > 0$ and $z$ near zero in $Σ_{η}$ with $z + h \in Σ_{η}$ , use the semigroup law and uniform bound to get $∥ \tilde{T} (z) x - x ∥ \leq (M_{0} + 1) ∥ x - T (h) x ∥ + ∥ \tilde{T} (z + h) x - T (h) x ∥ + ∥ T (h) x - x ∥$ . Continuity at the interior point $h$ makes the middle term tend to zero as $z \to 0$ ; strong continuity of $T$ at zero lets $∥ T (h) x - x ∥$ be arbitrarily small, so $\tilde{T}$ is strongly continuous at the vertex. Thus $\tilde{T}$ is a bounded analytic semigroup of angle $η > 0$ extending $T$ , and (a) holds.

3.2step 1.7step 2.1step 2.2L6givenalgebra

(a) implies (c) and (d). Assume (a); [step 2.1] gives (b) and [step 2.2] gives (e), so the contour semigroup of [step 1.7] is a bounded analytic semigroup generated by $A$ , and by the uniqueness clause of [L6] it equals the given $T$ ; the smoothing estimates of [L6] therefore give $T (t) X \subseteq D (A)$ and $\sup_{t > 0} t ∥ A T (t) ∥ \leq C_{1} < \infty$ , which is (c); moreover for $r > 0$ , $s \neq 0$ the point $λ = r + i s$ lies in $Σ_{π / 2 + ϑ - ε}$ and [step 2.2] gives $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣ \leq M_{ε} / ∣ s ∣$ , which is (d).

3.3step 1.7step 2.1step 2.2givenalgebra

The maximal angle. Let $ψ$ be the supremum of the $δ \in (0, π / 2]$ for which $A$ has a bounded analytic semigroup extension to $Σ_{δ}$ , let $B$ be the set of $ϑ \in (0, π / 2)$ for which (b) holds, and let $E$ be the set of $δ > 0$ with $A$ sectorial of exponent $δ$ ; [step 2.1] shows $(0, ψ) \subseteq B$ , so $\sup B \geq ψ$ , while $ϑ \in B$ gives $ϑ \in E$ by [step 2.2] and then an extension to $Σ_{ϑ}$ by [step 1.7], so $ϑ \leq ψ$ and $\sup B = ψ$ ; likewise $(0, ψ) \subseteq E$ by [step 2.2], so $\sup E \geq ψ$ , and $δ \in E$ gives an extension to $Σ_{δ}$ by [step 1.7], so $δ \leq ψ$ and $\sup E = ψ$ ; these are equalities of suprema only, with no attainment asserted at $ψ$ .

4.1step 1.7step 2.3step 3.1step 3.2step 3.3givenalgebra∎

Assembly. The implications (a) $\Rightarrow$ (b) [step 2.1], (b) $\Rightarrow$ (e) [step 2.2], (e) $\Rightarrow$ (a),(c) [step 1.7], (c) $\Rightarrow$ (a) [step 3.1], (a) $\Rightarrow$ (c),(d) [step 3.2] and (d) $\Rightarrow$ (e) [step 2.3] close the cycle, so (a)-(e) are equivalent; when they hold, the semigroup of (c) is the contour semigroup by the uniqueness argument of [step 3.2]; the angle equalities are [step 3.3]; and no choice principle beyond Dependent Choice is used in the argument: every inverse appearing is an explicit absolutely convergent Laplace integral, a norm limit of resolvents, or a Neumann series, boundedness of each inverse is proved explicitly, so the closed-graph implication in the resolvent vocabulary is not invoked.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Self-adjoint nonpositive operators generate bounded analytic semigroups

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 $H$ be a complex Hilbert space and let $A$ be a self-adjoint operator on $H$ with dense domain (Symmetric, self-adjoint and essentially self-adjoint operators, Unbounded linear operators: domain, graph and extension) satisfying the quadratic nonpositivity $⟨ A u, u ⟩ \leq 0$ for every $u \in D (A)$ . Then:

(1) for every $λ = a + i b$ with $b \neq 0$ , $∥ (λ I - A) u ∥ \geq ∣ b ∣ ∥ u ∥$ for all $u \in D (A)$ , $λ \in ρ (A)$ and $∥ R (λ, A) ∥ \leq 1 / ∣ b ∣$ ; more generally $∥ R (λ, A) ∥ \leq 1 / dist (λ, {z : Re z \leq 0})$ for $Re λ > 0$ ;

(2) $A$ is sectorial of angle $π / 2$ in the $e^{t A}$ convention (Sectorial operator with the semigroup sign convention) and generates a bounded analytic semigroup of angle $π / 2$ which is contractive on $[0, \infty)$ : $∥ T (t) ∥ \leq 1$ . Dependent Choice is assumed for the semigroup suppliers; Countable Choice is inherited from the vocabulary item Symmetric, self-adjoint and essentially self-adjoint operators; the proof below uses no choice principle beyond Dependent Choice.

Facts & Assumptions

Given: A complex Hilbert space $H$ , a densely defined self-adjoint operator $A$ on $H$ with $⟨ A u, u ⟩ \leq 0$ for all $u \in D (A)$ , and the numbers $a_{u} : = ⟨ A u, u ⟩ / ∥ u ∥_{H}^{2}$ for $u \in D (A) ∖ {0}$ .

[L1]

Self-adjointness means $T = T^{*}$ : domains and values agree, under Countable Choice for the adjoint vocabulary (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator).

[L2]

For a densely defined $T$ one has $ran (T - z)^{⊥} = \ker (T^{*} - \overline{z})$ for every $z \in C$ (The adjoint is well defined, closed, and reverses inclusions).

[L3]

The adjoint $T^{*}$ is closed (The adjoint is well defined, closed, and reverses inclusions).

[L4]

Cauchy-Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[L5]

$z \in ρ (T)$ when $z - T : D (T) \to H$ is bijective with bounded inverse $R_{T} (z) = (z - T)^{- 1}$ (Resolvent and spectrum of an unbounded operator).

[L6]

$A$ is sectorial of angle $δ > 0$ 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 $ε \in (0, δ)$ (Sectorial operator with the semigroup sign convention).

[L7]

The conditions (a)-(e) of the sectorial resolvent characterisation are equivalent, and when they hold the generated semigroup is the contour semigroup (Sectorial resolvent characterisation of bounded analytic semigroups).

[L8]

On a Hilbert space $A$ is dissipative if and only if $Re ⟨ A u, u ⟩ \leq 0$ for all $u \in D (A)$ , and Lumer-Phillips makes a densely defined dissipative $A$ generate a contraction semigroup if and only if $Ran (λ_{0} I - A) = H$ for some $λ_{0} > 0$ (Dissipative operator, Lumer-Phillips generation theorem).

[L9]

The contour semigroup generated by $A$ is the unique strongly continuous semigroup generated by $A$ within the class of exponentially bounded semigroups (The generator of the contour semigroup is the sectorial operator).

[L13]

Under Countable Choice, $N^{⊥ ⊥} = \overline{N}$ for every linear subspace $N$ of a Hilbert space (The double orthogonal complement of a subspace is its closure).

Proof

technique · direct

1.1L1L3L4givenalgebra

The lower bound, injectivity and closed range. For $u \in D (A) ∖ {0}$ put $a_{u} : = ⟨ A u, u ⟩ / ∥ u ∥_{H}^{2} \leq 0$ ; by [L4], $∥ (λ I - A) u ∥ ∥ u ∥_{H} \geq ∣ ⟨ (λ I - A) u, u ⟩ ∣ = ∣ λ - a_{u} ∣ ∥ u ∥_{H}^{2} \geq dist (λ, (- \infty, 0]) ∥ u ∥_{H}^{2}$ , so for every $λ \notin (- \infty, 0]$ the operator $λ I - A$ is injective with the lower bound $∥ (λ I - A) u ∥ \geq dist (λ, (- \infty, 0]) ∥ u ∥$ ; moreover its range is closed, because if $(λ I - A) u_{n} \to w$ then $(u_{n})$ is Cauchy, $u_{n} \to u$ and $A u_{n} = λ u_{n} - (λ I - A) u_{n} \to λ u - w$ , and closedness of $A$ from [L3] and [L1] gives $u \in D (A)$ with $(λ I - A) u = w$ .

2.1step 1.1L1L2L13givenalgebra

The range is dense. If $v ⊥ Ran (λ I - A)$ for some $λ \notin (- \infty, 0]$ , then [L2] with $T = A$ and $z = λ$ gives $v \in \ker (A^{*} - \overline{λ})$ ; by self-adjointness [L1] this says $A v = \overline{λ} v$ , that is $(\overline{λ} I - A) v = 0$ , and the lower bound of [step 1.1] at $\overline{λ}$ (which also lies outside $(- \infty, 0]$ ) forces $v = 0$ ; hence $Ran (λ I - A)^{⊥} = {0}$ and, since the range is closed by [step 1.1], [L13] gives $Ran (λ I - A) = {0}^{⊥} = H$ .

3.1step 1.1step 2.1L5L6givenalgebra

The resolvent bounds and sectoriality. By [step 2.1] and [step 1.1] the map $λ I - A$ is bijective with inverse bounded by $1 / dist (λ, (- \infty, 0])$ , so $λ \in ρ (A)$ with $∥ R (λ, A) ∥ \leq 1 / dist (λ, (- \infty, 0])$ for every $λ \notin (- \infty, 0]$ by [L5]; for $Re λ > 0$ the distance to the smaller set $(- \infty, 0]$ dominates the distance to ${z : Re z \leq 0}$ , which equals $Re λ$ , and for $λ = a + i b$ with $b \neq 0$ the distance to the real set $(- \infty, 0]$ is at least $∣ b ∣$ ; finally, for $∣ \arg λ ∣ \leq π - ε$ the nearest point of $(- \infty, 0]$ is the origin when $∣ \arg λ ∣ \leq π / 2$ and has distance $∣ λ ∣ \sin (π - ∣ \arg λ ∣) \geq ∣ λ ∣ \sin ε$ otherwise, so $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ with $M_{ε} = 1 / \sin ε$ on $Σ_{π - ε}$ ; since $Σ_{π} = C ∖ (- \infty, 0] \subseteq ρ (A)$ , this is sectoriality of angle $π / 2$ .

4.1step 3.1L1L7L8L9givenalgebra∎

Generation and contractivity. By [step 3.1] $A$ satisfies the sectorial resolvent condition with exponent $π / 2$ , so [L7] provides a bounded analytic semigroup $T$ of angle $π / 2$ generated by $A$ ; separately $A$ is dissipative by [L8] because $Re ⟨ A u, u ⟩ \leq 0$ , and $Ran (λ I - A) = H$ for every $λ > 0$ by [step 3.1], so Lumer-Phillips [L8] makes $A$ generate a strongly continuous contraction semigroup; that semigroup is bounded, hence exponentially bounded, and has generator $A$ , so by uniqueness [L9] it coincides with the analytic semigroup $T$ , giving $∥ T (t) ∥ \leq 1$ for $t \geq 0$ ; the argument assumes Dependent Choice and inherits Countable Choice from the adjoint vocabulary [L1] and uses no further choice principle beyond Dependent Choice.

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Quadratic spectral bounds control a self-adjoint parabolic 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 $H$ be a complex Hilbert space and let $A$ be a self-adjoint densely defined operator satisfying the quadratic upper bound $⟨ A u, u ⟩ \leq - λ_{1} ∥ u ∥^{2}$ for every $u \in D (A)$ and some $λ_{1} \in R$ (Symmetric, self-adjoint and essentially self-adjoint operators). Then $A$ generates a holomorphic semigroup family $(T (z))_{z \in Σ_{π / 2} \cup {0}}$ with $∥ T (z) ∥ \leq e^{- λ_{1} Re z}$ for every $z \in Σ_{π / 2}$ . If $λ_{1} \geq 0$ , this is a bounded analytic semigroup and $∥ T (z) ∥ \leq 1$ throughout the sector; if $λ_{1} > 0$ it decays exponentially on the positive real axis, $∥ T (t) ∥ \leq e^{- λ_{1} t}$ and $∥ T (t) x ∥ \leq e^{- λ_{1} t} ∥ x ∥$ for $t \geq 0$ . For $λ_{1} = 0$ the semigroup is contractive, while for $λ_{1} < 0$ the displayed estimate allows exponential growth. If $A$ is self-adjoint and $σ (A) \subseteq (- \infty, - λ_{1}]$ , then the quadratic hypothesis holds, so the same conclusion applies; passing from the spectral hypothesis to the quadratic one uses the projection-valued-measure spectral theorem and declares the Axiom of Choice exactly for that step (Spectral theorem for unbounded self-adjoint operators (PVM form)). The semigroup bound itself uses only the quadratic hypothesis.

Facts & Assumptions

Given: A complex Hilbert space $H$ , a self-adjoint densely defined operator $A$ with quadratic upper bound $⟨ A u, u ⟩ \leq - λ_{1} ∥ u ∥^{2}$ for a fixed real $λ_{1}$ and all $u \in D (A)$ , and the shifted operator $B : = A + λ_{1} I$ with $D (B) = D (A)$ .

[L1]

Self-adjointness means $T = T^{*}$ : domains and values agree; and $y \in D (T^{*})$ with $T^{*} y = w$ exactly when $⟨ T x, y ⟩ = ⟨ x, w ⟩$ for all $x \in D (T)$ (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator).

[L2]

A self-adjoint densely defined operator satisfying $⟨ A u, u ⟩ \leq 0$ is sectorial of angle $π / 2$ , generates a bounded analytic semigroup of angle $π / 2$ , and that semigroup is contractive on $[0, \infty)$ (Self-adjoint nonpositive operators generate bounded analytic semigroups).

[L3]

For $x \in D (A)$ the orbit of a strongly continuous semigroup is differentiable on $(0, \infty)$ with $\frac{d}{d t} T (t) x = T (t) A x = A T (t) x$ , and $T (t) x \in D (A)$ for all $t \geq 0$ (The generator commutes with the semigroup on its domain).

[L4]

On a star-shaped open set a continuous complex-differentiable $F$ has a holomorphic primitive $G$ with $G^{'} = F$ (Primitive and Cauchy theorem for Banach-valued holomorphic maps on star-shaped domains).

[L5]

A holomorphic Banach-space-valued function has norm-convergent power-series expansions; two power series about a real centre that agree on a real interval have equal coefficients (Cauchy integral formula and Cauchy estimates for Banach-valued holomorphic functions, Banach-valued power series are determined by their real values).

[L6]

The generator of a strongly continuous semigroup is closed, and for real $λ$ above the exponential growth bound it belongs to the resolvent set (The generator is closed and densely defined, Laplace transform formula for the resolvent, Resolvent and spectrum of a closed operator on a Banach space).

[L7]

On a Hilbert space an operator is dissipative exactly when $Re ⟨ A x, x ⟩ \leq 0$ for all $x \in D (A)$ (Dissipative operator).

[L8]

A bounded analytic semigroup of angle $δ$ is a family with $T (0) = I$ , the functional equation, operator-norm holomorphy on $Σ_{δ}$ , strong continuity at the vertex and uniform boundedness on smaller sectors (Complex sector and bounded analytic semigroup).

[L9]

The spectral theorem: a self-adjoint operator $A$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on the Borel sets of $R$ with $D (A) = {x : \int λ^{2} d E_{x} < \infty}$ and $A x = \int λ d E (λ) x$ ; the proof assumes the Axiom of Choice (Spectral theorem for unbounded self-adjoint operators (PVM form), The Axiom of Choice).

[L10]

For a PVM, $E (J)$ is an orthogonal projection and the scalar measure of $E (J) x$ is the restriction of $E_{x}$ to $J$ ; the PVM calculus gives $∥ (A - λ) y ∥^{2} = \int ∣ s - λ ∣^{2} d E_{y} (s)$ whenever $A = \int s d E$ and $y \in D (A)$ (Projection valued measure, Integral of a measurable function against a projection-valued measure, The unbounded PVM integral is densely defined, closed and normal, Bounded borel pvm integral).

Proof

technique · direct

1.1L1L2givenalgebra

The shifted operator. $D (B) = D (A)$ is dense; for $x, y \in D (A)$ one has $⟨ B x, y ⟩ = ⟨ A x, y ⟩ + λ_{1} ⟨ x, y ⟩ = ⟨ x, A y ⟩ + λ_{1} ⟨ x, y ⟩ = ⟨ x, B y ⟩$ , so $B$ is symmetric, and [L1] identifies $D (B^{*})$ with the $y$ for which $x \mapsto ⟨ A x, y ⟩$ is bounded on $D (A)$ , namely $D (A^{*}) = D (A)$ , with $B^{*} y = A^{*} y + λ_{1} y = B y$ ; hence $B$ is self-adjoint, and $⟨ B u, u ⟩ = ⟨ A u, u ⟩ + λ_{1} ∥ u ∥^{2} \leq 0$ for every $u \in D (A)$ ; by [L2] $B$ is sectorial of angle $π / 2$ and generates a bounded analytic semigroup $S$ of angle $π / 2$ that is contractive on $[0, \infty)$ .

2.1step 1.1L1L3L4L5L6L8givenalgebra

The rotated generators. Fix $α \in (- π / 2, π / 2)$ and put $V (t) : = S (e^{i α} t)$ for $t \geq 0$ : the functional equation of $S$ makes $V$ a semigroup, strong continuity at $0$ holds along the ray $e^{i α} [0, \infty) \subseteq Σ_{π / 2}$ by [L8], and $∥ V (t) ∥ \leq C_{α}$ for a finite constant because the ray lies in a strictly smaller sector; moreover for $f \in D (B)$ the identity $S (z) f - f = \int_{0}^{z} S (w) B f d w$ holds on $Σ_{π / 2}$ , because both sides are holomorphic by [L4] and [L5] (the primitive of $w \mapsto S (w) B f$ is holomorphic with derivative $S (z) B f$ ) and they agree on the real axis by the fundamental theorem and the orbit derivative $S^{'} (t) f = S (t) B f$ of [L3], so the Banach-valued identity theorem proved in Sectorial resolvent characterisation of bounded analytic semigroups extends the identity to the sector; dividing by $z = e^{i α} t$ and letting $t ↓ 0$ gives $(V (t) f - f) / t \to e^{i α} B f$ , so the generator $C_{α}$ of $V$ contains the closed operator $e^{i α} B$ ; for real $λ > 0$ one has $λ \in ρ (C_{α})$ by [L6] and $λ \in ρ (e^{i α} B)$ with $R (λ, e^{i α} B) = e^{- i α} R (e^{- i α} λ, B)$ because $e^{- i α} λ \notin (- \infty, 0]$ and $B$ is self-adjoint nonpositive; both operators are closed and their resolvents at $λ$ agree (the identity $(λ - C_{α}) R (λ, e^{i α} B) = I$ holds on $H$ since the two operators agree on $D (B)$ ), so [L1] and the resolvent definition give $C_{α} = e^{i α} B$ .

3.1step 2.1L3L7givenalgebra

Contractivity on the sector. The operator $e^{i α} B$ is dissipative by [L7], because for $v \in D (B)$ one has $Re ⟨ e^{i α} B v, v ⟩ = \cos α ⟨ B v, v ⟩ \leq 0$ (the value $⟨ B v, v ⟩ = \overline{⟨ v, B v ⟩}$ is real and nonpositive by self-adjointness and the quadratic bound); by [step 2.1] it is the generator of $V$ , so [L3] gives $\frac{d}{d t} ∥ V (t) f ∥^{2} = 2 Re ⟨ V (t) f, e^{i α} B V (t) f ⟩ \leq 0$ for $f \in D (B)$ and hence $∥ V (t) f ∥ \leq ∥ f ∥$ ; since $D (B)$ is dense and $V (t)$ is bounded this extends to all $f \in H$ , so $∥ S (z) ∥ \leq 1$ for every $z = e^{i α} t \in Σ_{π / 2}$ .

4.1step 1.1step 3.1L8givenalgebra

The semigroup generated by $A$ . Define $T (z) : = e^{- λ_{1} z} S (z)$ on $Σ_{π / 2} \cup {0}$ : the functional equation, operator-norm holomorphy and strong continuity at the vertex are inherited, and $∥ T (z) ∥ \leq e^{- λ_{1} Re z} ∥ S (z) ∥ \leq e^{- λ_{1} Re z}$ by [step 3.1]. For every real $λ_{1}$ this is a holomorphic semigroup family with generator $A$ , since for $f \in D (A) = D (B)$ the real difference quotient satisfies $\frac{e^{- λ_{1} t} S (t) f - f}{t} = \frac{S (t) f - f}{t} + \frac{e^{- λ_{1} t} - 1}{t} S (t) f \to B f - λ_{1} f = A f$ ; conversely $S (t) = e^{λ_{1} t} T (t)$ shows that a vector with a convergent $T$ difference quotient belongs to $D (B) = D (A)$ , so the generator is exactly $A$ . When $λ_{1} \geq 0$ the bound is at most $1$ on the whole sector, so $T$ is a bounded analytic semigroup of angle $π / 2$ ; when $λ_{1} > 0$ it gives the stated strict exponential decay on the real axis.

5.1step 4.1L9L10givenalgebra∎

The spectral clause. If $H = {0}$ , the conclusion is immediate. Otherwise assume AC and $σ (A) \subseteq (- \infty, - λ_{1}]$ , and let $E$ be supplied by [L9]. To prove its carrier assertion, fix a real $λ \in ρ (A)$ and $0 < r < ∥ R (λ, A) ∥^{- 1}$ . For $J = (λ - r, λ + r)$ , $y = E (J) x$ belongs to $D (A)$ because $J$ is bounded. By [L10], $∥ (A - λ) y ∥ \leq r ∥ y ∥$ , whereas the bounded inverse gives $∥ y ∥ \leq ∥ R (λ, A) ∥ ∥ (A - λ) y ∥$ . Hence $y = 0$ , and $E (J) = 0$ . Every real point outside $σ (A)$ has such a neighborhood; a countable rational-interval base gives a countable cover of this open set by subsets of these zero-projection neighborhoods. Countable additivity therefore gives $E (R ∖ σ (A)) = 0$ . Now [L9, L10] yield $⟨ A u, u ⟩ = \int s d E_{u} (s) \leq - λ_{1} E_{u} (R) = - λ_{1} ∥ u ∥^{2}$ for $u \in D (A)$ , so step 4.1 applies. AC is used in the spectral branch; the quadratic branch inherits Countable Choice from the self-adjoint generation supplier.

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The Dirichlet Laplacian generates an analytic 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.

Assume Countable Choice. For the bounded-domain compact-resolvent and eigenvalue-attainment clause in (2), additionally assume the Axiom of Choice.

Let $Ω \subseteq R^{n}$ be nonempty and open, $n \geq 1$ , and let $a_{0} (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H_{0}^{1} (Ω)$ be the principal Dirichlet form (The $L^{2}$ operator associated with a symmetric elliptic form, Zero-boundary Sobolev space as a norm closure). Let $A$ be the $L^{2} (Ω)$ operator associated with $a_{0}$ in the $e^{t A}$ convention, $D (A) = {u \in H_{0}^{1} (Ω) : \exists f \in L^{2} (Ω) with a_{0} (u, v) = - (f, v)_{L^{2}} \forall v \in H_{0}^{1} (Ω)}, A u = f .$ Then:

(1) $A$ is densely defined, self-adjoint and satisfies $⟨ A u, u ⟩ = - a_{0} (u, u) \leq 0$ ; hence $A = Δ_{D}$ and generates a contraction analytic semigroup $(T (t))_{t \geq 0}$ of maximal allowed angle $π / 2$ , with $∥ T (t) ∥ \leq 1$ ;

(2) if $a_{0}$ is coercive on $H_{0}^{1} (Ω)$ , the Rayleigh infimum $λ_{*} : = \inf_{0 \neq u \in H_{0}^{1} (Ω)} \frac{a_{0} (u, u)}{∥ u ∥_{2}^{2}}$ is positive and $∥ T (t) ∥ \leq e^{- λ_{*} t}$ . A sufficient condition for coercivity is that $Ω$ lie in a slab of finite width, by zero extension and the one-dimensional Poincare inequality. If in addition $Ω$ is bounded and the Axiom of Choice holds, compact resolvent makes $λ_{*}$ an attained first Dirichlet eigenvalue; for unbounded $Ω$ , the infimum need not be attained.

(3) If $Ω$ is a bounded $C^{2}$ domain, then $D (A) = H^{2} (Ω) \cap H_{0}^{1} (Ω)$ and $A u = Δ u$ for $u \in D (A)$ , with graph norm equivalent to the $H^{2}$ norm (Global $H^{2}$ Dirichlet regularity);

(4) for $m \geq 1$ the domain $D (A^{m})$ is the recursive graph domain ${u \in D (A^{m - 1}) : A u \in D (A^{m - 1})}$ and $T (t) L^{2} \subseteq D (A^{m})$ for $t > 0$ ; if $Ω$ is bounded $C^{2 m}$ , then $D (A^{m}) = {u \in H^{2 m} (Ω) : Δ^{j} u \in H_{0}^{1} (Ω), 0 \leq j < m}$ , but no such spatial identification may be asserted without the boundary compatibility (Higher-order boundary regularity for Dirichlet problems). Dependent Choice is assumed throughout for the semigroup suppliers; Countable Choice is assumed for self-adjointness, and the elliptic-regularity suppliers. The Axiom of Choice is used additionally only for the bounded-domain compact-resolvent and eigenvalue-attainment clause in (2); the unbounded-domain generation claim in (1) is preserved.

Facts & Assumptions

Given: Countable Choice; and, only when the bounded-domain compact-resolvent/eigenvalue-attainment clause is invoked, the Axiom of Choice; a nonempty open set $Ω \subseteq R^{n}$ with $n \geq 1$ ; the complex Hilbert space $L^{2} (Ω)$ with its inner product $(\cdot, \cdot)_{L^{2}}$ ; the principal Dirichlet form $a_{0} (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H_{0}^{1} (Ω)$ , i.e. the symmetric divergence-form case with $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ ; the Rayleigh infimum $λ_{*} : = \inf {a_{0} (u, u) / ∥ u ∥_{L^{2}}^{2} : 0 \neq u \in H_{0}^{1} (Ω)}$ ; the operator $A$ of the statement, defined by the weak identity $a_{0} (u, v) = - (A u, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ ; the form operator $L$ of [L1] with $a_{0} (u, v) = (L u, v)_{L^{2}}$ ; the iterated graph domains $D (A^{0}) : = L^{2} (Ω)$ and $D (A^{m}) : = {u \in D (A^{m - 1}) : A u \in D (A^{m - 1})}$ for $m \geq 1$ ; and the semigroup $T$ of assertion (1).

[L1]

The symmetric-case operator of The $L^{2}$ operator associated with a symmetric elliptic form on the Hilbert space $L^{2} (Ω)$ (Hilbert space) is $D (L) = {u \in H_{0}^{1} (Ω) : \exists f \in L^{2} (Ω) with a_{0} (u, v) = (f, v)_{L^{2}} \forall v \in H_{0}^{1} (Ω)}$ with $L u : = f$ , well defined and linear, and $H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ is the closure of $C_{c}^{\infty} (Ω)$ in $W^{1, 2} (Ω)$ (Zero-boundary Sobolev space as a norm closure).

[L2]

In this symmetric case with $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ , the domain $D (L)$ is dense in $L^{2} (Ω)$ , $L$ is symmetric, and $(L u, u)_{L^{2}} = a_{0} (u, u) \geq 0$ for every $u \in D (L)$ (The associated elliptic operator is densely defined, symmetric and lower bounded).

[L3]

In the complex scalar-field case $K = C$ of The symmetric elliptic form operator is self-adjoint with compact resolvent, Countable Choice gives $L = L^{*}$ on $L^{2} (Ω; C)$ ; its real-case complexification branch is not needed here.

[L4]

For a densely defined linear operator $T$ the adjoint $T^{*}$ is well defined with $D (T^{*}) = {y : x \mapsto ⟨ T x, y ⟩ bounded}$ and $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ for $x \in D (T)$ , $y \in D (T^{*})$ ; $T^{*}$ is closed and reverses inclusions (Adjoint of a densely defined operator, The adjoint is well defined, closed, and reverses inclusions).

[L5]

A self-adjoint densely defined operator $A$ with $⟨ A u, u ⟩ \leq 0$ for every $u \in D (A)$ is sectorial of angle $π / 2$ with vertex $0$ in the $e^{t A}$ convention (Sectorial operator with the semigroup sign convention) and generates a bounded analytic semigroup of angle $π / 2$ (Complex sector and bounded analytic semigroup) which is contractive on $[0, \infty)$ (Self-adjoint nonpositive operators generate bounded analytic semigroups).

[L6]

A bounded analytic semigroup of angle $δ \in (0, π / 2]$ is a family $(T (z))_{z \in Σ_{δ} \cup {0}}$ with $T (0) = I$ , the functional equation, operator-norm holomorphy on $Σ_{δ}$ , strong continuity at the vertex and uniform boundedness on every strictly smaller sector; its generator is the generator of $(T (t))_{t \geq 0}$ , and its angle is the supremum of the $δ$ for which such a family exists and extends the given one (Complex sector and bounded analytic semigroup).

[L7]

A self-adjoint densely defined $A$ with $⟨ A u, u ⟩ \leq - λ ∥ u ∥^{2}$ for real $λ$ generates a holomorphic semigroup family with $∥ T (z) ∥ \leq e^{- λ Re z}$ on $Σ_{π / 2}$ , in particular $∥ T (t) ∥ \leq e^{- λ t}$ for $t \geq 0$ ; when $λ \geq 0$ the family is a bounded analytic semigroup of angle $π / 2$ , and when $λ > 0$ it decays exponentially (Quadratic spectral bounds control a self-adjoint parabolic semigroup).

[L8]

In the sign convention of this track the symbol $Δ_{D}$ denotes the $L^{2}$ operator associated with the Dirichlet energy form for the equation $u^{'} = A u$ , i.e. the operator whose weak identity reads $a_{0} (u, v) = - (A u, v)_{L^{2}}$ ; it is not $- Δ_{D}$ that generates the heat flow (Semigroup sign and generator conventions).

[L9]

If $Ω$ is open and there are a unit vector $e$ and $a < b$ with $a < x \cdot e < b$ for every $x \in Ω$ , then $∥ u ∥_{L^{2} (Ω)} \leq C (2) (b - a) ∥ D u ∥_{L^{2} (Ω)}$ for every $u \in W_{0}^{1, 2} (Ω)$ , with a positive constant $C (2)$ ; Countable Choice is assumed (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[L10]

If $Ω$ is nonempty bounded open, then the first Dirichlet eigenvalue satisfies $λ_{1} > 0$ , $λ_{1} = \min_{0 \neq u \in H_{0}^{1} (Ω)} ∥ D u ∥_{L^{2}}^{2} / ∥ u ∥_{L^{2}}^{2}$ with the minimum attained exactly at the nonzero first eigenfunctions, and $λ_{1}$ is the smallest eigenvalue of an orthonormal eigenbasis $(e_{j})_{j \geq 1} \subseteq H_{0}^{1} (Ω)$ of $L^{2} (Ω)$ with $λ_{j} \to + \infty$ ; these suppliers assume the Axiom of Choice and Countable Choice (The Poincare constant is the reciprocal square root of the first Dirichlet eigenvalue, The Rayleigh principle for the first Dirichlet eigenvalue, Discrete spectrum of a symmetric elliptic Dirichlet operator, The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L11]

On an interval $I = (0, L)$ one has $∥ u ∥_{L^{2} (I)} \leq (L / π) ∥ u^{'} ∥_{L^{2} (I)}$ for every $u \in H_{0}^{1} (I)$ , with equality for $ϕ (x) = \sin (π x / L)$ , and $\int_{I} ϕ^{'} \overline{v^{'}} d x = (π / L)^{2} \int_{I} ϕ \overline{v} d x$ for every $v \in H_{0}^{1} (I)$ (The sharp Dirichlet Poincare inequality on an interval).

[L12]

Global $H^{2}$ Dirichlet regularity, under Countable Choice: for a bounded $C^{2}$ domain $Ω \subseteq R^{n}$ with $n \geq 2$ , if $u \in H_{0}^{1} (Ω)$ weakly solves $L u = f$ with $f \in L^{2} (Ω)$ , then $u \in H^{2} (Ω)$ and $∥ u ∥_{H^{2} (Ω)} \leq C (∥ f ∥_{L^{2} (Ω)} + ∥ u ∥_{L^{2} (Ω)})$ (Global $H^{2}$ Dirichlet regularity).

[L13]

Higher-order boundary regularity, under Countable Choice: for a bounded $C^{k + 2}$ domain $Ω \subseteq R^{n}$ with $n \geq 2$ , if $u \in H_{0}^{1} (Ω)$ weakly solves $L u = f$ with $f \in H^{k} (Ω)$ , then $u \in H^{k + 2} (Ω)$ with $∥ u ∥_{H^{k + 2} (Ω)} \leq C (∥ f ∥_{H^{k} (Ω)} + ∥ u ∥_{L^{2} (Ω)})$ (Higher-order boundary regularity for Dirichlet problems).

[L14]

In dimension $n = 1$ the bounded sets satisfying the local one-sided condition of Bounded C^k domains and boundary charts are finite disjoint unions of bounded open intervals.

[L15]

For a sectorial operator $A$ of angle $δ \in (0, π / 2]$ with vertex $0$ , 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 generator $A$ (Smoothing estimates for the semigroup generated by a sectorial operator, The generator of the contour semigroup is the sectorial operator).

[L16]

The regularity conclusions of [L12] and [L13] assume that the given weak solution already belongs to $H_{0}^{1} (Ω)$ and that the domain and forcing have the stated regularity. Thus they do not by themselves prove existence or remove these hypotheses (Global $H^{2}$ Dirichlet regularity, Higher-order boundary regularity for Dirichlet problems).

[L17]

If $Ω$ is bounded and the Axiom of Choice holds, then the shifted solution operator $K_{μ}$ is compact, so $L$ has compact resolvent; the Axiom of Choice is used additionally only for this bounded-domain compactness clause (The symmetric elliptic form operator is self-adjoint with compact resolvent).

[L18]

$W^{k, p}$ membership means that for every multi-index $α$ with $∣ α ∣ \leq k$ there is an $L^{p}$ class $D^{α} u$ satisfying the weak-derivative identity against every test function, and the test pairing is bilinear, without conjugation (Integer-order Sobolev spaces and their norms).

Proof

technique · direct

1.1L1L2givenalgebra

The operator $A$ is well defined with $D (A) = D (L)$ and $A u = - L u$ : by [L1], the condition $a_{0} (u, v) = - (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ says exactly that the pair $(u, g)$ with $g : = - f$ satisfies $a_{0} (u, v) = (g, v)_{L^{2}}$ for all $v$ , which is the defining identity of membership in $D (L)$ with $L u = g$ ; conversely, if $u \in D (L)$ with datum $g$ , then $a_{0} (u, v) = (g, v)_{L^{2}} = (- (- g), v)_{L^{2}}$ for all $v$ , so $u \in D (A)$ with the datum $f = - g$ ; hence $u \in D (A)$ if and only if $u \in D (L)$ and $A u = - L u$ , the datum being unique in both definitions, so $D (A)$ is dense in $L^{2} (Ω)$ by [L2] and $⟨ A u, u ⟩ = - (L u, u)_{L^{2}} = - a_{0} (u, u) \leq 0$ for every $u \in D (A)$ again by [L2].

1.2L1L9givenalgebra

The slab criterion. If there are a unit vector $e$ and $a < b$ with $a < x \cdot e < b$ for every $x \in Ω$ , then [L9] at $p = 2$ gives $∥ u ∥_{L^{2} (Ω)} \leq C (2) (b - a) ∥ D u ∥_{L^{2} (Ω)}$ for every $u \in H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ by [L1], hence $a_{0} (u, u) = ∥ D u ∥_{L^{2}}^{2} \geq (C (2) (b - a))^{- 2} ∥ u ∥_{L^{2}}^{2}$ , and $∥ u ∥_{H^{1}}^{2} \leq (1 + C (2)^{2} (b - a)^{2}) ∥ D u ∥_{2}^{2}$ , so $a_{0}$ is coercive on $H_{0}^{1}$ ; this is the zero-extension, one-coordinate-dimension reduction performed by [L9], applied directly on $Ω$ .

1.3L10L17given

The bounded case under Axiom of Choice. Assume $Ω$ is bounded and the Axiom of Choice holds; then the resolvent of $L$ is compact by [L17], and [L10] supplies $λ_{1} > 0$ with $λ_{1} = \min_{0 \neq u \in H_{0}^{1} (Ω)} ∥ D u ∥_{L^{2}}^{2} / ∥ u ∥_{L^{2}}^{2}$ attained exactly at the nonzero first eigenfunctions and with $λ_{1}$ the smallest element of the eigenvalue list of an orthonormal eigenbasis $(e_{j}) \subseteq H_{0}^{1} (Ω)$ of $L^{2} (Ω)$ , so $λ_{*} = λ_{1}$ is positive and attained by a first eigenfunction; Countable Choice enters through [L10], and Axiom of Choice is used for the compact-resolvent/spectral branch through [L17].

1.4L1L11givenalgebra

Unbounded domains need not attain. Fix $L > 0$ , put $L_{k} : = L k / (k + 1)$ , $I_{k} : = (k (L + 1), k (L + 1) + L_{k})$ and $Ω : = ⋃_{k \geq 1} I_{k}$ ; since $L_{k} < L$ , successive intervals have gap $(L + 1) - L_{k} > 1$ , so the components are disjoint. Every $u \in H_{0}^{1} (Ω)$ satisfies $\int_{Ω} ∣ u ∣^{2} = \sum_{k} \int_{I_{k}} ∣ u ∣^{2}$ and $\int_{Ω} ∣ u^{'} ∣^{2} = \sum_{k} \int_{I_{k}} ∣ u^{'} ∣^{2}$ , and the restriction $u ∣_{I_{k}}$ lies in $H_{0}^{1} (I_{k})$ , because restriction to a component carries $C_{c}^{\infty} (Ω)$ into $C_{c}^{\infty} (I_{k})$ with classical derivatives restricting to classical derivatives and is bounded for the $W^{1, 2}$ norm, hence carries the closure of the test functions into the closure; on each $I_{k}$ the sharp inequality [L11] gives $\int_{I_{k}} ∣ u^{'} ∣^{2} \geq (π / L_{k})^{2} \int_{I_{k}} ∣ u ∣^{2} \geq (π / L)^{2} \int_{I_{k}} ∣ u ∣^{2}$ because $L_{k} < L$ , so $a_{0} (u, u) \geq (π / L)^{2} ∥ u ∥_{L^{2}}^{2}$ and $λ_{*} \geq (π / L)^{2}$ , while the functions $ϕ_{k} (x) : = \sin (π (x - k (L + 1)) / L_{k})$ on $I_{k}$ , extended by zero, lie in $H_{0}^{1} (Ω)$ because they lie in $H_{0}^{1} (I_{k})$ by [L11] and are limits of compactly supported smooth functions on $I_{k}$ zero-extended to $Ω$ , and have Rayleigh quotient $(π / L_{k})^{2}$ , so $λ_{*} \leq \inf_{k} (π / L_{k})^{2} = (π / L)^{2}$ ; hence $λ_{*} = (π / L)^{2}$ is positive and $a_{0}$ is coercive, but the infimum is not attained: otherwise $\sum_{k} \int_{I_{k}} (∣ u^{'} ∣^{2} - λ_{*} ∣ u ∣^{2}) = 0$ with every summand at least $((π / L_{k})^{2} - (π / L)^{2}) \int_{I_{k}} ∣ u ∣^{2} \geq 0$ and every coefficient strictly positive, forcing $\int_{I_{k}} ∣ u ∣^{2} = 0$ for all $k$ and hence $u = 0$ , which is excluded from the Rayleigh quotient.

2.1step 1.1L3L4algebra

Self-adjointness of $A$ . By [L3] $L = L^{*}$ with $D (L^{*}) = D (L)$ ; since $D (A) = D (L)$ and $A = - L$ by [step 1.1], the adjoint description of [L4] gives $D (A^{*}) = D ((- L)^{*}) = {y : x \mapsto ⟨ - L x, y ⟩ bounded} = {y : x \mapsto ⟨ L x, y ⟩ bounded} = D (L^{*}) = D (L) = D (A)$ and $⟨ x, A^{*} y ⟩ = ⟨ A x, y ⟩ = - ⟨ L x, y ⟩ = - ⟨ x, L y ⟩ = ⟨ x, - L y ⟩$ for all $x \in D (A)$ , so $A^{*} y = - L y = A y$ for every $y \in D (A)$ and $A$ is self-adjoint; with [step 1.1] $A$ is densely defined and satisfies $⟨ A u, u ⟩ \leq 0$ .

2.2step 1.1L8given

Identification with the Dirichlet Laplacian. By [L8] the symbol $Δ_{D}$ denotes the $L^{2}$ operator associated with the Dirichlet energy form in the $u^{'} = A u$ convention, whose defining weak identity is exactly $a_{0} (u, v) = - (A u, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ ; [step 1.1] shows that $A$ has this defining identity, so $A = Δ_{D}$ .

2.3step 1.1L1L12L18algebra

Domain identification for $n \geq 2$ . Assume $n \geq 2$ , $Ω$ a bounded $C^{2}$ domain, and let $u \in D (A) = D (L)$ with datum $f = L u \in L^{2} (Ω)$ ; [L12] applied to the weak equation $L u = f$ gives $u \in H^{2} (Ω)$ with $∥ u ∥_{H^{2} (Ω)} \leq C (∥ L u ∥_{L^{2}} + ∥ u ∥_{L^{2}}) = C (∥ A u ∥_{L^{2}} + ∥ u ∥_{L^{2}})$ ; conversely, if $u \in H^{2} (Ω) \cap H_{0}^{1} (Ω)$ , then for every $φ \in C_{c}^{\infty} (Ω)$ the weak-derivative identities of [L18] applied to $\partial_{j} u \in H^{1} (Ω)$ with test function $\overline{φ}$ give $a_{0} (u, φ) = \sum_{j} \int_{Ω} \partial_{j} u \overline{\partial_{j} φ} = - \sum_{j} \int_{Ω} \partial_{j}^{2} u \overline{φ} = - (Δ u, φ)_{L^{2}}$ , and because $C_{c}^{\infty} (Ω)$ is dense in $H_{0}^{1} (Ω)$ by [L1] while $v \mapsto a_{0} (u, v)$ and $v \mapsto (Δ u, v)_{L^{2}}$ are both continuous on $H_{0}^{1} (Ω)$ , the identity $a_{0} (u, v) = - (Δ u, v)_{L^{2}}$ holds for every $v \in H_{0}^{1} (Ω)$ ; hence $u \in D (L)$ with $L u = - Δ u$ , so $u \in D (A)$ with $A u = Δ u$ ; therefore $D (A) = H^{2} (Ω) \cap H_{0}^{1} (Ω)$ and $A u = Δ u$ on $D (A)$ , and the estimate above together with $∥ Δ u ∥_{L^{2}} \leq n ∥ u ∥_{H^{2} (Ω)}$ shows that the graph norm $∥ u ∥_{L^{2}} + ∥ A u ∥_{L^{2}}$ is equivalent to $∥ u ∥_{H^{2} (Ω)}$ .

2.4step 1.1L1L14L18givenalgebra

Domain identification for $n = 1$ . Assume $n = 1$ ; by [L14] a bounded $C^{2}$ domain is then a finite disjoint union of bounded open intervals $I_{k}$ , and the following argument applies to each component and each $u \in D (A) = D (L)$ : for $φ \in C_{c}^{\infty} (I_{k})$ , extended by zero to $Ω$ , the identity $a_{0} (u, φ) = (f, φ)_{L^{2}}$ with datum $f = L u$ reads $\int_{I_{k}} u^{'} \overline{φ^{'}} = \int_{I_{k}} f \overline{φ}$ , and replacing $φ$ by $\overline{φ}$ in the bilinear weak-derivative convention of [L18] shows that $- f$ is the weak derivative of the class $u^{'} \in L^{2} (I_{k})$ ; hence $u^{'}$ has an $L^{2}$ weak derivative, $u \in W^{2, 2} (I_{k}) = H^{2} (I_{k})$ , and $L u = f = - u^{''}$ ; conversely, if $u \in H^{2} (Ω) \cap H_{0}^{1} (Ω)$ , then $u^{''} \in L^{2} (Ω)$ and the same weak-derivative identity gives $a_{0} (u, φ) = (- u^{''}, φ)_{L^{2}}$ for every $φ \in C_{c}^{\infty} (Ω)$ , which extends to every $v \in H_{0}^{1} (Ω)$ by density by [L1]; hence $u \in D (A)$ with $A u = Δ u = u^{''}$ ; consequently $D (A) = H^{2} (Ω) \cap H_{0}^{1} (Ω)$ and $A u = Δ u$ in dimension one as well, and the graph norm is equivalent to the $H^{2}$ norm, since $∥ u^{'} ∥_{L^{2}}^{2} = - (u^{''}, u)_{L^{2}} \leq ∥ u^{''} ∥_{L^{2}} ∥ u ∥_{L^{2}}$ by [step 1.1] gives $∥ u ∥_{H^{2} (Ω)}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ u^{'} ∥_{L^{2}}^{2} + ∥ u^{''} ∥_{L^{2}}^{2} \leq (∥ u ∥_{L^{2}} + ∥ u^{''} ∥_{L^{2}})^{2} \leq 2 (∥ u ∥_{L^{2}} + ∥ A u ∥_{L^{2}})^{2}$ while $∥ u ∥_{L^{2}} + ∥ A u ∥_{L^{2}} \leq 2 ∥ u ∥_{H^{2} (Ω)}$ .

3.1step 2.1L5L6algebra

Generation and maximal angle. Applying [L5] to the self-adjoint operator $A$ of [step 2.1] with its quadratic bound of [step 1.1] yields that $A$ is sectorial of angle $π / 2$ with vertex $0$ in the $e^{t A}$ convention and generates a bounded analytic semigroup $(T (z))_{z \in Σ_{π / 2} \cup {0}}$ of angle $π / 2$ with generator $A$ , contractive on $[0, \infty)$ ; by the definition of the analytic-semigroup angle in [L6] the admissible angles are the $δ \in (0, π / 2]$ for which such a family exists, so the exhibited family realizes the maximal allowed angle $π / 2$ , and $∥ T (t) ∥ \leq 1$ for $t \geq 0$ .

3.2step 2.3L13givenalgebra

Higher-order domains for $n \geq 2$ . Assume $n \geq 2$ and $Ω$ a bounded $C^{2 m}$ domain, and let $E_{m} : = {u \in H^{2 m} (Ω) : Δ^{j} u \in H_{0}^{1} (Ω), 0 \leq j < m}$ ; the claim is $D (A^{m}) = E_{m}$ for every $m \geq 1$ , and $E_{1} = D (A)$ with $A u = Δ u$ is [step 2.3]; if $u \in D (A^{m})$ with $m \geq 2$ , then $u \in D (A^{m - 1})$ and $A u \in D (A^{m - 1})$ , so the induction hypothesis gives $u \in H^{2 m - 2} (Ω)$ with $Δ^{j} u \in H_{0}^{1} (Ω)$ for $j < m - 1$ , and gives $A u \in H^{2 m - 2} (Ω)$ with $Δ^{j} (A u) \in H_{0}^{1} (Ω)$ for $j < m - 1$ ; since $A u = Δ u$ by [step 2.3], the datum $f = L u = - A u$ lies in $H^{2 m - 2} (Ω)$ , so [L13] with $k = 2 m - 2$ applied to the weak solution $u \in H_{0}^{1} (Ω)$ upgrades $u$ to $H^{2 m} (Ω)$ , and $Δ^{j} u \in H_{0}^{1} (Ω)$ for $j < m$ holds by the induction hypothesis and $Δ^{j} (A u) \in H_{0}^{1} (Ω)$ ; hence $D (A^{m}) \subseteq E_{m}$ ; conversely, if $u \in E_{m}$ with $m \geq 2$ , then $u \in E_{1} = D (A)$ with $A u = Δ u$ , and $Δ u \in H^{2 m - 2} (Ω)$ has $Δ^{j} (Δ u) = Δ^{j + 1} u \in H_{0}^{1} (Ω)$ for $j < m - 1$ , so $Δ u \in E_{m - 1} = D (A^{m - 1})$ by the induction hypothesis, and $u \in D (A)$ with $A u = Δ u \in D (A^{m - 1})$ gives $u \in D (A^{m})$ by the recursive definition; hence $E_{m} \subseteq D (A^{m})$ and $D (A^{m}) = E_{m}$ for all $m \geq 1$ .

3.3step 2.4L14L18algebra

Higher-order domains for $n = 1$ . If $n = 1$ , then $Ω$ is a finite disjoint union of bounded open intervals by [L14], and the weak-derivative argument of [step 2.4] gives, by induction on $m$ , the description $D (A^{m}) = {u \in H^{2 m} (Ω) : Δ^{j} u \in H_{0}^{1} (Ω), 0 \leq j < m}$ : the case $m = 1$ is [step 2.4]; if $u \in D (A^{m})$ with $m \geq 2$ , then $u \in D (A^{m - 1})$ and $A u \in D (A^{m - 1})$ , so the induction hypothesis applied to $u$ gives $u \in H^{2 m - 2} (Ω)$ with $u^{(2 j)} \in H_{0}^{1} (Ω)$ for $j < m - 1$ and applied to $A u = u^{''}$ gives $u^{''} \in H^{2 m - 2} (Ω)$ with $u^{(2 j + 2)} \in H_{0}^{1} (Ω)$ for $j < m - 1$ , hence $u \in H^{2 m} (Ω)$ and all $u^{(2 j)}$ , $0 \leq j < m$ , lie in $H_{0}^{1} (Ω)$ ; conversely, if $u \in H^{2 m} (Ω)$ has $u^{(2 j)} \in H_{0}^{1} (Ω)$ for $j < m$ , then $u \in D (A)$ with $A u = u^{''}$ by [step 2.4] and $u^{''} \in H^{2 m - 2} (Ω)$ satisfies $(u^{''})^{(2 j)} = u^{(2 j + 2)} \in H_{0}^{1} (Ω)$ for $j < m - 1$ , so $u^{''} \in D (A^{m - 1})$ by the induction hypothesis and $u \in D (A^{m})$ ; in dimension one no boundary-regularity hypothesis is needed, because the weak-derivative identity of [L18] is available on every open set.

4.1step 1.1step 3.1L7L15givenalgebra

Exponential decay under coercivity. If $a_{0} (u, u) \geq α ∥ u ∥_{L^{2}}^{2}$ for some $α > 0$ and all $u \in H_{0}^{1} (Ω)$ , then $λ_{*} \geq α > 0$ , and [step 1.1] gives $⟨ A u, u ⟩ = - a_{0} (u, u) \leq - λ_{*} ∥ u ∥^{2}$ for every $u \in D (A)$ , so the $λ_{*} > 0$ branch of [L7] produces a bounded analytic semigroup of angle $π / 2$ generated by $A$ with $∥ T (z) ∥ \leq e^{- λ_{*} Re z}$ on $Σ_{π / 2}$ ; by [L15] the contour semigroup is the unique exponentially bounded strongly continuous semigroup with generator $A$ , and both the family of [L7] and the family $T$ of [step 3.1] are exponentially bounded and have generator $A$ , so they coincide and $∥ T (t) ∥ \leq e^{- λ_{*} t}$ for $t \geq 0$ .

4.2step 3.1L15algebra

Positive-time smoothing. By [L15] applied to the sectorial operator $A$ of [step 3.1], the semigroup $T$ generated by $A$ satisfies $T (t) L^{2} (Ω) \subseteq D (A^{m})$ for every $t > 0$ and $m \geq 1$ , where the graph domains are the recursive domains of the statement: $D (A^{m}) = {u \in D (A^{m - 1}) : A u \in D (A^{m - 1})}$ is equivalent to ${u \in D (A^{m - 1}) : A^{m - 1} u \in D (A)}$ by induction on $m$ , and $D (A^{0}) = L^{2} (Ω)$ .

5.1step 3.2step 3.3step 4.2L9L16givenalgebra∎

The boundary-compatibility caveat. The spatial identifications in steps 3.2 and 3.3 retain $Δ^{j} u \in H_{0}^{1} (Ω)$ for every $0 \leq j < m$ : these conditions follow from the recursive operator domain and cannot be discarded merely because $u \in H^{2 m} (Ω)$ . For example, on a nonempty bounded smooth domain the constant function $1$ belongs to every $H^{2 m}$ but not to $H_{0}^{1}$ , since [L9] would give $∥ 1 ∥_{2} \leq C ∥ \nabla 1 ∥_{2} = 0$ if it did. Moreover [L16] shows that the regularity suppliers for $n \geq 2$ apply only under their boundary and data hypotheses. Thus the abstract smoothing of step 4.2 supplies the recursive graph domain on general open sets; the stated spatial conclusions require their additional hypotheses. This reasoning uses neither a trace lifting nor any additional choice assumption.

Remarks

The generation asserted in (1) uses the quadratic bound of [step 1.1] and [L5], with Countable Choice entering through the self-adjointness supplier [L3]; the elliptic estimates [L12] and [L13] also assume Countable Choice. Axiom of Choice is used additionally only in the bounded-domain compact-resolvent/spectral branch [step 1.3] through [L10] and [L17]; no step uses any further choice principle. The maximal angle $π / 2$ of (1) is the largest angle the definition of a bounded analytic semigroup admits, and no claim is made that the semigroup is bounded on any larger sector, nor that the unshifted semigroup of a general coercive form is bounded without the self-adjointness used here.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Closed sectorial form and its associated operator

Definition

Let $H$ be a complex Hilbert space with inner product $(\cdot, \cdot)$ linear in the first argument and conjugate-linear in the second (Hilbert space, Real and complex inner-product spaces and their induced length), and let $V \subseteq H$ be a dense linear subspace carrying a Hilbert norm $∥ \cdot ∥_{V}$ whose inclusion $V ↪ H$ is continuous (The Sobolev space $H^{1}$ is a Hilbert space; the standard instance is $V = H_{0}^{1} (Ω) \subseteq H = L^{2} (Ω)$ ). A sesquilinear form $a : V \times V \to C$ , linear in the first argument and conjugate-linear in the second (Bounded, coercive and symmetric sesquilinear forms), is a closed sectorial form on $V \subseteq H$ if:

(i) $a$ is bounded on $V$ : there is $C < \infty$ with $∣ a (u, v) ∣ \leq C ∥ u ∥_{V} ∥ v ∥_{V}$ for all $u, v \in V$ ;

(ii) there are $M \geq 0$ and $θ \in [0, π / 2)$ with $Re a (u, u) \geq - M ∥ u ∥_{H}^{2}, ∣ Im a (u, u) ∣ \leq \tan θ (Re a (u, u) + M ∥ u ∥_{H}^{2})$ for every $u \in V$ ;

(iii) $V$ is complete for the shifted form norm $∥ u ∥_{a, M}^{2} : = Re a (u, u) + M ∥ u ∥_{H}^{2} + ∥ u ∥_{H}^{2},$ By (ii) the square is nonnegative and $∥ u ∥_{H} \leq ∥ u ∥_{a, M}$ , so $∥ u ∥_{a, M} = 0$ only for $u = 0$ ; condition (iii) is the closedness of the form. The Hermitian pairing $b (u, v) = (a (u, v) + \overline{a (v, u)}) / 2 + (M + 1) (u, v)_{H}$ has $b (u, u) = ∥ u ∥_{a, M}^{2} > 0$ for $u \neq 0$ , so it is an inner product and this expression is its norm (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

Comparison with a prescribed form-domain norm. Under Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain), $∥ \cdot ∥_{a, M}$ is equivalent to $∥ \cdot ∥_{V}$ . The upper bound is $∥ u ∥_{a, M}^{2} \leq (C + (M + 1) ∥ ι ∥^{2}) ∥ u ∥_{V}^{2}$ . The identity in the reverse direction has closed graph: convergence in either norm implies convergence in $H$ , so the two limits agree. Both normed versions of $V$ are Banach, and Closed graph theorem makes this inverse identity bounded, giving a positive lower comparison constant. Generation results below may instead take this norm equivalence as an explicit hypothesis, retaining only Countable Choice for Lax–Milgram.

The associated operator $A$ of such a form, in the $e^{t A}$ sign convention, is $D (A) : = {u \in V : \exists f \in H with a (u, v) = - (f, v) for all v \in V}, A u : = f .$

The vector $f$ is unique, so $A$ is well defined: if $f, g$ both satisfy the relation, then $(f - g, v) = 0$ for every $v \in V$ ; since $V$ is dense in $H$ and the inner product is continuous, $(f - g, w) = 0$ for every $w \in H$ , and testing $w = f - g$ gives $∥ f - g ∥^{2} = 0$ . The operator $A$ is linear (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores): if $u_{1}, u_{2} \in D (A)$ with associated vectors $f_{1}, f_{2}$ and $λ \in C$ , then $a (u_{1} + λ u_{2}, v) = a (u_{1}, v) + λ a (u_{2}, v) = - (f_{1} + λ f_{2}, v)$ for all $v \in V$ , so $u_{1} + λ u_{2} \in D (A)$ with $A (u_{1} + λ u_{2}) = f_{1} + λ f_{2}$ . The defining relation reads $⟨ A u, v ⟩ = - a (u, v) (u \in D (A), v \in V),$ and is the only sign convention used on this page.

The form is coercive with constant $α > 0$ when $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ for all $u \in V$ . A coercive form is closed: (iii) holds with explicit estimates, since (i) and coercivity give, with $ι : V ↪ H$ the inclusion, $α ∥ u ∥_{V}^{2} \leq ∥ u ∥_{a, M}^{2} \leq (C + (M + 1) ∥ ι ∥^{2}) ∥ u ∥_{V}^{2} .$

Sign convention

The dictionary $⟨ A u, v ⟩ = - a (u, v)$ fixes the orientation: for $u \in D (A)$ the quadratic form of $A$ is the negative of $a$ , $⟨ A u, u ⟩ = - a (u, u),$ so the sector condition (ii) says that the shifted operator $M - A$ is accretive, $Re ⟨ (M - A) u, u ⟩ = Re a (u, u) + M ∥ u ∥_{H}^{2} \geq 0,$ and gives $Re ⟨ (A - M) u, u ⟩ \leq 0$ for every $u \in D (A)$ . This is the convention in which $e^{t A}$ solves $u^{'} = A u$ and the resolvent sector of $A$ opens to the right; the opposite pairing $a (u, v) = ⟨ A u, v ⟩$ is the Pazy-Lunardi convention for $- A$ and is not used here.

Coercivity versus sectoriality

A coercive form satisfies (ii) with $M = 0$ and $θ = \arctan (C / α)$ at most, because $Re a (u, u) \geq α ∥ u ∥_{V}^{2} \geq 0$ and $∣ Im a (u, u) ∣ \leq C ∥ u ∥_{V}^{2}$ . The sectorial condition is a one-sided quantitative hypothesis on $Re a$ and $Im a$ along the diagonal; the phrase "elliptic operator" is never used as a hypothesis (Real and imaginary parts, complex conjugation, and modulus).

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The sectorial form angle controls the numerical range of its operator

Statement

Let $a$ be a closed sectorial form on $V \subseteq H$ with constants $M, θ$ and associated operator $A$ (Closed sectorial form and its associated operator), and write $\overline{S_{θ}} : = {ζ \in C : ∣ \arg ζ ∣ \leq θ} \cup {0}$ for the closed sector of half-angle $θ$ around the positive real axis. Then for every $u \in D (A)$ :

$⟨ A u, u ⟩ = - a (u, u)$ ;
$⟨ (A - M) u, u ⟩ \in - \overline{S_{θ}}$ .

In particular the normalized quadratic form values $⟨ (A - M) u, u ⟩ / ∥ u ∥_{H}^{2}$ for $u \in D (A) ∖ {0}$ lie in $- \overline{S_{θ}}$ and those of $A$ lie in its translate by $M$ ; when $A$ is bounded this is the containment of the numerical range (Numerical range and numerical radius). No choice principle is used.

Facts & Assumptions

Given: A closed sectorial form $a$ on the dense subspace $V \subseteq H$ with constants $M \geq 0$ and $θ \in [0, π / 2)$ and associated operator $A$ (Closed sectorial form and its associated operator); the closed sector $\overline{S_{θ}} = {ζ : Re ζ \geq 0, ∣ Im ζ ∣ \leq \tan θ Re ζ}$ ; and a vector $u \in D (A)$ with $f : = A u$ .

[L1]

$D (A) = {u \in V : \exists f \in H, a (u, v) = - (f, v) \forall v \in V}$ and $A u : = f$ , and the form satisfies $Re a (u, u) \geq - M ∥ u ∥_{H}^{2}$ and $∣ Im a (u, u) ∣ \leq \tan θ (Re a (u, u) + M ∥ u ∥_{H}^{2})$ for all $u \in V$ ; the pairing is linear in the first argument (Closed sectorial form and its associated operator).

[L2]

The inner product of a complex Hilbert space satisfies $⟨ u, v ⟩ = \overline{⟨ v, u ⟩}$ , $⟨ v, v ⟩ \geq 0$ with equality only for $v = 0$ , and is linear in the first argument; in particular $⟨ v, v ⟩ = ∥ v ∥_{H}^{2}$ is real and nonnegative (Real and complex inner-product spaces and their induced length, Hilbert space).

Proof

technique · direct

1.1L1givenalgebra

Claim 1. For $u \in D (A)$ the defining relation with $f = A u$ gives $a (u, v) = - (A u, v)$ for every $v \in V$ ; testing with $v = u$ gives $a (u, u) = - (A u, u)$ , that is $⟨ A u, u ⟩ = - a (u, u)$ .

1.2L1L2givenalgebra

The shifted form and its sector. Define $b (u, v) : = a (u, v) + M ⟨ u, v ⟩$ on $V \times V$ . Then $b$ is sesquilinear and, for $u \in V$ , $Re b (u, u) = Re a (u, u) + M ∥ u ∥_{H}^{2} \geq 0$ by [L1] and [L2], while $Im b (u, u) = Im a (u, u)$ and $∣ Im a (u, u) ∣ \leq \tan θ (Re a (u, u) + M ∥ u ∥_{H}^{2})$ , so $∣ Im b (u, u) ∣ \leq \tan θ Re b (u, u)$ ; by the description of $\overline{S_{θ}}$ in the givens this says $b (u, u) \in \overline{S_{θ}}$ .

2.1step 1.1step 1.2L2algebra

Claim 2. For $u \in D (A)$ , $⟨ (A - M) u, u ⟩ = ⟨ A u, u ⟩ - M ⟨ u, u ⟩ = - a (u, u) - M ∥ u ∥_{H}^{2} = - (a (u, u) + M ⟨ u, u ⟩) = - b (u, u)$ by [step 1.1], [step 1.2] and [L2], and $- b (u, u) \in - \overline{S_{θ}}$ by [step 1.2].

3.1step 1.1step 2.1L2given∎

Normalized consequences. If $u \in D (A) ∖ {0}$ then $∥ u ∥_{H}^{2} > 0$ by [L2], and multiplying by the positive real scalar $∥ u ∥_{H}^{- 2}$ preserves the closed sector $\overline{S_{θ}}$ , so $⟨ (A - M) u, u ⟩ / ∥ u ∥_{H}^{2} \in - \overline{S_{θ}}$ and $⟨ A u, u ⟩ / ∥ u ∥_{H}^{2} \in M - \overline{S_{θ}}$ ; for bounded $A$ on a nonzero $H$ , these normalized values are exactly the numerical ranges of $A - M$ and $A$ , respectively. On $H = {0}$ both operators are zero and their numerical ranges are ${0}$ by the convention in Numerical range and numerical radius; the containments still hold since $0 \in - \overline{S_{θ}}$ and $0 = M - M \in M - \overline{S_{θ}}$ . Claims 1 and 2 are [step 1.1] and [step 2.1], and the argument fixed the arbitrary vector $u$ and used no selection, so no choice principle was used.

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Coercive sectorial forms define closed densely defined sectorial operators

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 $a$ be a closed sectorial form on $V \subseteq H$ with constants $M, θ$ and associated operator $A$ (Closed sectorial form and its associated operator). Choose any $κ > 0$ satisfying the continuous embedding bound $∥ u ∥_{H} \leq κ ∥ u ∥_{V}$ for every $u \in V$ ; such a positive bound exists, including when $H = {0}$ . By the comparison clause of Closed sectorial form and its associated operator under Dependent Choice, the closed form norm $∥ u ∥_{a, M}^{2} : = Re a (u, u) + (M + 1) ∥ u ∥_{H}^{2}$ is equivalent to $∥ u ∥_{V}^{2}$ ; fix $c_{a} > 0$ such that $∥ u ∥_{a, M}^{2} \geq c_{a} ∥ u ∥_{V}^{2}$ . Then:

(1) $A$ is closed and $D (A)$ is dense in $H$ ;

(2) for every $λ$ with $Re λ > M$ , the form $a_{λ} (u, v) : = a (u, v) + λ (u, v)$ is coercive on $V$ with constant $β_{λ} : = c_{a} \min {1, Re λ - M} > 0$ . The Lax-Milgram solution of $a (u, v) + λ (u, v) = (f, v)$ is the unique $u \in D (A)$ with $(λ I - A) u = f$ , and satisfies $∥ u ∥_{V} \leq κ β_{λ}^{- 1} ∥ f ∥_{H}, ∥ u ∥_{H} \leq κ^{2} β_{λ}^{- 1} ∥ f ∥_{H};$

(3) $A$ is sectorial with vertex $M$ and every exponent $δ < π / 2 - θ$ in the sense of Sectorial operator with the semigroup sign convention; in particular, $∥ R (λ, A) ∥ \leq K_{ε} / ∣ λ - M ∣$ on $M + Σ_{π / 2 + δ - ε}$ ;

(4) $B : = A - M I$ generates a bounded analytic semigroup $S$ on every sector $Σ_{δ}$ with $δ < π / 2 - θ$ , and $T (z) : = e^{M z} S (z)$ is the analytic semigroup generated by $A$ , with $∥ T (z) ∥ \leq C_{δ^{'}} e^{M Re z}$ on each smaller sector $Σ_{δ^{'}}$ with $δ^{'} < δ$ . If $a$ is coercive with $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ , then $∥ T (t) ∥ \leq e^{- α t / κ^{2}}$ for $t \geq 0$ . Countable Choice is inherited from The Lax--Milgram theorem; no additional choice principle is used.

Facts & Assumptions

Given: A closed sectorial form $a$ on $V \subseteq H$ with constants $M \geq 0$ , $θ \in [0, π / 2)$ , associated operator $A$ , and a positive embedding bound $κ > 0$ with $∥ u ∥_{H} \leq κ ∥ u ∥_{V}$ for all $u \in V$ ; a form bound $C$ on $V$ ; and a constant $c_{a} > 0$ with $∥ u ∥_{a, M}^{2} : = Re a (u, u) + (M + 1) ∥ u ∥_{H}^{2} \geq c_{a} ∥ u ∥_{V}^{2}$ .

[L1]

The associated operator is $D (A) = {u \in V : \exists f \in H with a (u, v) = - (f, v) \forall v \in V}$ and $A u = f$ (Closed sectorial form and its associated operator).

[L2]

$V \subseteq H$ is a dense linear subspace carrying a Hilbert norm $∥ \cdot ∥_{V}$ whose inclusion into $H$ is continuous, and the chosen positive constant $κ$ satisfies $∥ u ∥_{H} \leq κ ∥ u ∥_{V}$ ; the shifted form norm satisfies $∥ u ∥_{a, M}^{2} = Re a (u, u) + M ∥ u ∥_{H}^{2} + ∥ u ∥_{H}^{2}$ (Closed sectorial form and its associated operator).

[L3]

A form is coercive with constant $α > 0$ when $Re a (u, u) \geq α ∥ u ∥^{2}$ for all $u$ (Bounded, coercive and symmetric sesquilinear forms).

[L4]

Lax-Milgram assumes Countable Choice, and for a bounded coercive form on a Hilbert space with coercivity constant $α$ and a bounded conjugate-linear functional $F$ it produces a unique $u$ with $a (u, v) = F (v)$ for all $v$ and the estimate $∥ u ∥ \leq ∥ F ∥ / α$ (The Lax--Milgram theorem, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L5]

For $u \in D (A)$ one has $⟨ A u, u ⟩ = - a (u, u)$ (The sectorial form angle controls the numerical range of its operator).

[L6]

For $u \in D (A)$ one has $⟨ (A - M) u, u ⟩ \in - \overline{S_{θ}}$ , where $\overline{S_{θ}}$ is the closed sector of half-angle $θ$ around the positive real axis (The sectorial form angle controls the numerical range of its operator).

[L7]

Cauchy-Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[L8]

The resolvent set is open: if $λ_{0} \in ρ (A)$ and $∣ λ - λ_{0} ∣ ∥ R (λ_{0}, A) ∥ < 1$ then $λ \in ρ (A)$ (Resolvent identity and holomorphy for a closed operator, Resolvent and spectrum of a closed operator on a Banach space).

[L9]

The conditions (a)-(e) of the sectorial resolvent characterisation are equivalent; in particular condition (e) holds exactly when $A$ has a bounded analytic semigroup extension and generates a bounded strongly continuous semigroup (Sectorial resolvent characterisation of bounded analytic semigroups).

[L12]

If the equivalent conditions hold, the generated semigroup is the contour semigroup (Sectorial resolvent characterisation of bounded analytic semigroups).

[L10]

[L11]

On a Hilbert space an operator is dissipative if and only if $Re ⟨ A x, x ⟩ \leq 0$ for every $x \in D (A)$ ; and Lumer-Phillips: a densely defined dissipative operator generates a strongly continuous contraction semigroup if and only if $Ran (λ_{0} I - A) = X$ for some $λ_{0} > 0$ (Dissipative operator, Lumer-Phillips generation theorem).

[L13]

Under Countable Choice, $N^{⊥ ⊥} = \overline{N}$ for every linear subspace $N$ of a Hilbert space (The double orthogonal complement of a subspace is its closure).

Proof

technique · direct

1.1L1L2L3L4givenalgebra

Coercivity and the resolvent solution. Put $q (u) : = Re a (u, u) + M ∥ u ∥_{H}^{2} \geq 0$ , so that $∥ u ∥_{a, M}^{2} = q (u) + ∥ u ∥_{H}^{2}$ and, for $Re λ > M$ , $Re a_{λ} (u, u) = q (u) + (Re λ - M) ∥ u ∥_{H}^{2} \geq \min {1, Re λ - M} ∥ u ∥_{a, M}^{2} \geq β_{λ} ∥ u ∥_{V}^{2}$ , while $∣ a_{λ} (u, v) ∣ \leq (C + ∣ λ ∣ κ^{2}) ∥ u ∥_{V} ∥ v ∥_{V}$ and the functional $F (v) : = (f, v)_{H}$ is conjugate-linear with $∥ F ∥ \leq κ ∥ f ∥_{H}$ ; Lax-Milgram on the Hilbert space $V$ therefore gives, for each $f \in H$ , a unique $u \in V$ with $a_{λ} (u, v) = (f, v)_{H}$ for all $v \in V$ and the bounds $∥ u ∥_{V} \leq κ β_{λ}^{- 1} ∥ f ∥_{H}$ and $∥ u ∥_{H} \leq κ^{2} β_{λ}^{- 1} ∥ f ∥_{H}$ , and rewriting the weak equation as $a (u, v) = - (λ u - f, v)$ for all $v$ gives $u \in D (A)$ with $A u = λ u - f$ , that is $(λ I - A) u = f$ ; conversely every $u \in D (A)$ with $(λ I - A) u = f$ satisfies the weak equation and is therefore this unique solution.

2.1step 1.1L1givenalgebra

Closedness of $A$ . Fix a real $λ > M$ ; by [step 1.1] the map $λ I - A : D (A) \to H$ is bijective with $∥ u ∥_{H} \leq κ^{2} β_{λ}^{- 1} ∥ (λ I - A) u ∥_{H}$ , so if $u_{n} \in D (A)$ with $u_{n} \to u$ and $A u_{n} \to g$ then $λ u_{n} - A u_{n} \to λ u - g$ and $u_{n} = R (λ, A) (λ u_{n} - A u_{n}) \to R (λ, A) (λ u - g)$ ; comparing limits gives $u = R (λ, A) (λ u - g) \in D (A)$ and $(λ I - A) u = λ u - g$ , that is $A u = g$ , so the graph of $A$ is closed.

2.2step 1.1L1L2L5L13givenalgebra

Density of $D (A)$ . Let $f \in H$ satisfy $(f, w) = 0$ for every $w \in D (A)$ ; by [step 1.1] with a real $λ > M$ there is $u \in D (A)$ with $(λ I - A) u = f$ , so $(f, u) = 0$ while, by [L5], $Re (f, u) = λ ∥ u ∥_{H}^{2} + Re a (u, u) = q (u) + (λ - M) ∥ u ∥_{H}^{2} \geq (λ - M) ∥ u ∥_{H}^{2} \geq 0$ ; hence $u = 0$ and then $(f, v) = a_{λ} (u, v) = 0$ for every $v \in V$ ; since $V$ is dense in $H$ by [L2], continuity of the inner product gives $f = 0$ , so $D (A)^{⊥} = {0}$ and [L13] gives $\overline{D (A)} = H$ .

3.1step 1.1step 2.1L6L7L8givenalgebra

Sectoriality with vertex $M$ . For $u \in D (A) ∖ {0}$ the normalised value $z_{u} : = ⟨ A u, u ⟩ / ∥ u ∥_{H}^{2}$ lies in the closed sector $M - \overline{S_{θ}}$ by [L6], so for $λ$ outside $M - \overline{S_{θ}}$ one has $dist (λ, M - \overline{S_{θ}}) > 0$ and, by [L7], $∥ (λ I - A) u ∥ ∥ u ∥_{H} \geq ∣ ⟨ (λ I - A) u, u ⟩ ∣ = ∣ λ - z_{u} ∣ ∥ u ∥_{H}^{2} \geq dist (λ, M - \overline{S_{θ}}) ∥ u ∥_{H}^{2}$ ; thus $λ I - A$ is injective and $∥ R (λ, A) g ∥_{H} \leq ∥ g ∥_{H} / dist (λ, M - \overline{S_{θ}})$ on every surjectivity point $λ$ . Let $Ω : = C ∖ (M - \overline{S_{θ}})$ and $S : = {λ \in Ω : λ I - A is surjective}$ : $S$ is nonempty because $(M, \infty) \subseteq S$ by [step 1.1]; $S$ is open in $Ω$ because at $λ_{0} \in S$ injectivity and surjectivity give $∥ R (λ_{0}, A) ∥ \leq 1 / dist (λ_{0}, M - \overline{S_{θ}})$ and [L8] applies; $S$ is closed in $Ω$ because for $λ_{n} \in S$ with $λ_{n} \to λ \in Ω$ the resolvent identity gives $∥ R (λ_{n}, A) f - R (λ_{m}, A) f ∥ \leq ∣ λ_{n} - λ_{m} ∣ (2 / dist (λ, M - \overline{S_{θ}}))^{2} ∥ f ∥$ , so $u_{n} : = R (λ_{n}, A) f$ converges to some $u$ , and $λ_{n} u_{n} - A u_{n} = f$ with closedness of $A$ from [step 2.1] gives $u \in D (A)$ and $(λ I - A) u = f$ ; since $Ω$ is connected (the complement of a closed sector of opening angle $2 θ < π$ ), $S = Ω$ ; finally, for $δ < π / 2 - θ$ and $λ = M + r e^{i α}$ with $∣ α ∣ \leq π / 2 + δ - ε$ the angular distance from $λ$ to $M - \overline{S_{θ}}$ is at least $π / 2 - θ - δ + ε$ , so $dist (λ, M - \overline{S_{θ}}) \geq ∣ λ - M ∣ \cos (θ + δ - ε)$ and hence $∥ R (λ, A) ∥ \leq K_{ε} / ∣ λ - M ∣$ with $K_{ε} : = 1 / \cos (θ + δ - ε)$ on $M + Σ_{π / 2 + δ - ε}$ , which is sectoriality of vertex $M$ and every exponent $δ < π / 2 - θ$ .

4.1step 2.1step 2.2step 3.1L9L12L10givenalgebra

The shifted operator and the generated semigroup. Since $D (B) = D (A)$ is dense and $B$ is closed by [step 2.1] and [step 2.2], and since $λ I - B = (λ + M) I - A$ shows that $B$ is sectorial with vertex $0$ and every exponent $δ < π / 2 - θ$ together with the same $K_{ε} / ∣ λ ∣$ bound, the characterisation theorem [L9, L12] provides a bounded analytic semigroup $S$ of angle $δ$ generated by $B = A - M I$ on every such sector; the family $T (z) : = e^{M z} S (z)$ satisfies $T (0) = I$ , $T (z + w) = T (z) T (w)$ , is norm-holomorphic and strongly continuous, is bounded by $C_{δ^{'}} e^{M Re z}$ on each $Σ_{δ^{'}}$ , and has generator $A$ because for $x \in D (A)$ the difference quotient tends to $B x + M x = A x$ ; conversely $S (t) = e^{- M t} T (t)$ shows that a convergent $T$ difference quotient implies a convergent $S$ difference quotient, so the generator domain is exactly $D (A)$ ; it is therefore the analytic semigroup generated by $A$ , unique among exponentially bounded semigroups by [L10].

5.1step 1.1step 2.2step 4.1L5L10L11givenalgebra∎

The coercive case. If $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ , then for $u \in D (A)$ one has $Re ⟨ A u, u ⟩ = - Re a (u, u) \leq - α ∥ u ∥_{V}^{2} \leq - α κ^{- 2} ∥ u ∥_{H}^{2}$ by [L5] and $∥ u ∥_{H} \leq κ ∥ u ∥_{V}$ , so $A_{0} : = A + α κ^{- 2} I$ is densely defined by [step 2.2] and dissipative by [L11], and $Ran (λ_{0} I - A_{0}) = Ran ((λ_{0} - α κ^{- 2}) I - A) = H$ for $λ_{0} : = M + α κ^{- 2} + 1$ because $λ_{0} - α κ^{- 2} = M + 1 > M$ lies in $ρ (A)$ by [step 1.1]; hence $A_{0}$ generates a contraction semigroup $S^{'}$ by [L11], and the family $t \mapsto e^{- α κ^{- 2} t} S^{'} (t)$ is an exponentially bounded strongly continuous semigroup with generator $A$ , so it equals $T$ by [L10] and $∥ T (t) ∥ \leq e^{- α t / κ^{2}}$ for $t \geq 0$ ; the argument assumes Dependent Choice and inherits Countable Choice from the Lax-Milgram step [step 1.1] and uses no further choice principle beyond Dependent Choice.

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Form-generated sectorial elliptic semigroups

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.

The declared Dependent Choice assumption supplies equivalence of the shifted form norm with the prescribed norm on $V$ , by Closed sectorial form and its associated operator.

Let $H$ and $V$ be complex Hilbert spaces with $V \subseteq H$ dense and continuously embedded. Choose any $κ > 0$ satisfying $∥ v ∥_{H} \leq κ ∥ v ∥_{V}$ for every $v \in V$ ; when $H = {0}$ any positive $κ$ is admissible. Let $a$ be a closed sectorial form on $V$ with lower-bound constant $M \geq 0$ and sector half-angle $θ \in [0, π / 2)$ , with associated operator $A$ (Closed sectorial form and its associated operator). Then $B : = A - M I$ is sectorial with vertex $0$ and every exponent $δ < π / 2 - θ$ . It generates a bounded analytic semigroup $S$ on each such $Σ_{δ}$ , and $T (z) : = e^{M z} S (z)$ is the analytic semigroup generated by $A$ , satisfying $∥ T (z) ∥ \leq C_{δ^{'}} e^{M Re z}$ on every smaller sector $Σ_{δ^{'}}$ with $δ^{'} < δ$ . Thus the form assumptions guarantee analyticity on every sector strictly narrower than $π / 2 - θ$ ; they do not assert boundedness of the unshifted semigroup or that this lower angle is maximal. If the form is coercive with constant $α > 0$ , then $∥ T (t) ∥ \leq e^{- α t / κ^{2}}$ . In particular, for the symmetric Dirichlet form on a nonempty open $Ω$ the associated operator is $Δ_{D}$ and the abstract heat flow of The Dirichlet Laplacian generates an analytic heat semigroup is recovered. No symmetry is assumed. Dependent Choice is assumed for the semigroup suppliers; Countable Choice is inherited from the Lax-Milgram step; sectorial generation uses the declared Dependent Choice assumption.

Facts & Assumptions

Given: Complex Hilbert spaces $V \subseteq H$ with dense continuous inclusion and a chosen positive embedding bound $κ > 0$ satisfying $∥ v ∥_{H} \leq κ ∥ v ∥_{V}$ ; a closed sectorial form $a$ on $V$ with lower-bound constant $M \geq 0$ and sector half-angle $θ \in [0, π / 2)$ in the sense of [L2], with associated operator $A$ defined by $⟨ A u, v ⟩ = - a (u, v)$ for all $v \in V$ ; the shifted operator $B : = A - M I$ ; and, in the coercive clause, a constant $α > 0$ with $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ for all $u \in V$ .

[L1]

For a closed sectorial form with constants $M, θ$ and associated operator $A$ , using the chosen positive embedding bound $κ$ : $A$ is closed and $D (A)$ is dense in $H$ ; $A$ is sectorial with vertex $M$ and every exponent $δ < π / 2 - θ$ ; $B = A - M I$ generates a bounded analytic semigroup $S$ on every sector $Σ_{δ}$ with $δ < π / 2 - θ$ ; and $T (z) = e^{M z} S (z)$ is the analytic semigroup generated by $A$ , with $∥ T (z) ∥ \leq C_{δ^{'}} e^{M Re z}$ on each $Σ_{δ^{'}}$ with $δ^{'} < δ$ . Dependent Choice is assumed for the semigroup suppliers; Countable Choice is inherited from the Lax-Milgram step (Coercive sectorial forms define closed densely defined sectorial operators, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

A closed sectorial form on $V \subseteq H$ is a bounded sesquilinear form admitting $M \geq 0$ , $θ \in [0, π / 2)$ with $Re a (u, u) \geq - M ∥ u ∥_{H}^{2}$ and $∣ Im a (u, u) ∣ \leq \tan θ (Re a (u, u) + M ∥ u ∥_{H}^{2})$ , and complete for the shifted form norm; its equivalence to the prescribed $V$ norm follows under the declared Dependent Choice assumption; its associated operator is defined by $⟨ A u, v ⟩ = - a (u, v)$ for every $v \in V$ , and coercivity means $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ (Closed sectorial form and its associated operator).

[L3]

A bounded analytic semigroup of angle $δ$ is a strongly continuous-in-the-vertex, operator-norm holomorphic family on $Σ_{δ}$ satisfying the functional equation and bounded on every strictly smaller sector; its angle is the supremum of the admissible $δ$ (Complex sector and bounded analytic semigroup).

[L4]

For the principal Dirichlet form $a_{0}$ on a nonempty open $Ω$ , the associated 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 it generates a contraction analytic semigroup of maximal allowed angle $π / 2$ (The Dirichlet Laplacian generates an analytic heat semigroup).

[L5]

The contour semigroup is the unique strongly continuous semigroup with a given sectorial generator $A$ within the class of exponentially bounded semigroups (The generator of the contour semigroup is the sectorial operator).

[L6]

If $a$ is coercive with constant $α > 0$ and $κ > 0$ satisfies the chosen embedding bound $∥ v ∥_{H} \leq κ ∥ v ∥_{V}$ , then $∥ T (t) ∥ \leq e^{- α t / κ^{2}}$ for $t \geq 0$ (Coercive sectorial forms define closed densely defined sectorial operators).

Proof

technique · direct

1.1L1L2L3givenalgebra

The form-generated semigroup. By [L1], applied to the given closed sectorial form with constants $M, θ$ , the associated operator $A$ is closed and densely defined, is sectorial with vertex $M$ and every exponent $δ < π / 2 - θ$ , and $B = A - M I$ generates a bounded analytic semigroup $S$ on every $Σ_{δ}$ with $δ < π / 2 - θ$ ; moreover $B$ is sectorial with vertex $0$ and the same exponents, because $λ I - B = (λ + M) I - A$ for every $λ$ , so $λ \in ρ (B)$ exactly when $λ + M \in ρ (A)$ with $R (λ, B) = R (λ + M, A)$ and the defining bound $∥ R (λ + M, A) ∥ \leq K_{ε} / ∣ λ + M - M ∣$ becomes $∥ R (λ, B) ∥ \leq K_{ε} / ∣ λ ∣$ ; finally $T (z) = e^{M z} S (z)$ is the analytic semigroup generated by $A$ with $∥ T (z) ∥ \leq C_{δ^{'}} e^{M Re z}$ on every smaller sector $Σ_{δ^{'}}$ , $δ^{'} < δ$ , by [L1] and the definition of the analytic-semigroup angle in [L3].

1.2L6L2given

The coercive case. If $Re a (u, u) \geq α ∥ u ∥_{V}^{2}$ with $α > 0$ , then [L6] gives $∥ T (t) ∥ \leq e^{- α t / κ^{2}}$ for every $t \geq 0$ using the chosen positive embedding bound. If $H = {0}$ , density forces $V = {0}$ and the semigroup has norm $0$ , so the same estimate holds directly.

1.3L1L2L3given

The angle caveat. The assertion is that for every $δ < π / 2 - θ$ the shifted family $S$ is analytic and bounded on the sector $Σ_{δ}$ , and on each strictly smaller $Σ_{δ^{'}}$ the bound carries the factor $e^{M Re z}$ ; the definition [L3] asserts no family on $Σ_{π / 2 - θ}$ itself and makes no maximality claim, and when $M > 0$ the factor $e^{M Re z}$ is unbounded on any sector, so the unshifted semigroup $T$ is not asserted to be bounded; only the shifted semigroup $S$ is bounded, and no symmetry of the form is assumed, so none of the stronger conclusions of [L4] applies to a general $a$ .

2.1step 1.1L2L4L5givenalgebra∎

The Dirichlet specialisation. For a nonempty open $Ω$ , the principal form $a_{0} (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ is a closed sectorial form on $V = H_{0}^{1} (Ω) \subseteq H = L^{2} (Ω)$ with $M = 0$ and $θ = 0$ : it is bounded on $V$ by Cauchy-Schwarz, its real part is $∥ D u ∥_{L^{2}}^{2} \geq 0$ with vanishing imaginary part, and the shifted form norm is the complete $H^{1}$ norm; the associated operator of [L2] is exactly the operator $A$ of [L4], since both are defined by $a_{0} (u, v) = - (A u, v)_{L^{2}}$ ; the abstract construction of [step 1.1] therefore produces a bounded analytic semigroup generated by this $A$ , and [L4] identifies $A = Δ_{D}$ and exhibits the contraction analytic heat semigroup generated by it; by [L5] the semigroup constructed here and the heat semigroup of [L4] are the same exponentially bounded semigroup with generator $A$ , so the abstract heat flow is recovered, and no elliptic regularity or domain identification beyond [L4] is used.

Remarks

Dependent Choice is assumed for the semigroup suppliers; Countable Choice is inherited from the Lax-Milgram step of [L1] and from the vocabulary of [L2]; the rescalings and Euler exponentials of steps 1.1-2.1 use no choice principle. The theorem is a consolidation of the closed-form resolvent lemma [L1] with the Dirichlet specialisation of [L4]; no spatial domain identification is asserted for a general form, that role being reserved for the elliptic-regularity results cited in [L4].

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Analytic Duhamel cancellation removes the generator singularity

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 $A$ be sectorial of angle $δ \in (0, π / 2]$ in the $e^{t A}$ convention (Sectorial operator with the semigroup sign convention) and let $(T (t))_{t \geq 0}$ be the generated analytic semigroup with $∥ T (t) ∥ \leq c_{0}$ and $∥ A T (t) ∥ \leq c_{1} t^{- 1}$ for $0 < t \leq b$ (Smoothing estimates for the semigroup generated by a sectorial operator). Let $f \in C^{α} ([0, b], X)$ with Hölder constant $[f]_{α} < \infty$ , where $α \in (0, 1)$ . For $0 < t \leq b$ set $v (t) : = \int_{0}^{t} T (t - s) f (s) d s, v_{1} (t) : = \int_{0}^{t} T (t - s) (f (s) - f (t)) d s, v_{2} (t) : = \int_{0}^{t} T (τ) f (t) d τ,$ so that $v = v_{1} + v_{2}$ . Then:

$v_{1} (t) \in D (A)$ and $A v_{1} (t) = \int_{0}^{t} A T (t - s) (f (s) - f (t)) d s$ with $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ ;
$v_{2} (t) \in D (A)$ , $A v_{2} (t) = (T (t) - I) f (t)$ and $∥ A v_{2} (t) ∥ \leq (c_{0} + 1) ∥ f ∥_{\infty}$ ;
$v \in C ([0, b], D (A))$ in the graph norm and $A v \in C ([0, b], X)$ when $v (0) : = 0$ , with $A v (0) = 0$ .

Consequently, for $x \in D (A)$ the function $u (t) : = T (t) x + v (t)$ satisfies $u (t) \in D (A)$ and $A u (t) = A T (t) x + A v (t) \to A x$ as $t ↓ 0$ . No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with its analytic semigroup $T$ , constants $∥ T (t) ∥ \leq c_{0}$ , $∥ A T (t) ∥ \leq c_{1} / t$ on $(0, b]$ , a Hölder-continuous $f \in C^{α} ([0, b], X)$ with constant $[f]_{α}$ , an exponent $α \in (0, 1)$ , and the functions $v, v_{1}, v_{2}$ above; $f$ is continuous and hence Bochner integrable on $[0, b]$ , and $C^{α}$ embeds in $C ([0, b], X)$ .

[L1]

$T (t) X \subseteq D (A)$ for $t > 0$ , $A T (t) \in B (X)$ with $∥ A T (t) ∥ \leq c_{1} / t$ , and $∥ T (t) ∥ \leq c_{0}$ (Smoothing estimates for the semigroup generated by a sectorial operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L2]

For every $y \in X$ and $t > 0$ one has $J_{t} y : = \int_{0}^{t} T (τ) y d τ \in D (A)$ with $A J_{t} y = T (t) y - y$ ; and for $y \in D (A)$ one has $A T (t) y = T (t) A y$ (Time integrals of semigroup orbits lie in the generator domain, The generator commutes with the semigroup on its domain).

[L3]

The Bochner integral obeys $∥ \int_{E} g ∥ \leq \int_{E} ∥ g ∥$ and the Duhamel integral $t \mapsto \int_{0}^{t} T (t - s) f (s) d s$ is continuous on $[0, b]$ for continuous $f$ (Bochner integral norm inequality, The variation-of-constants integral is continuous for integrable forcing).

Proof

technique · direct

1.1L1L2L3givenalgebra

Truncated first term. Fix $0 < t \leq b$ and $0 < ε < t$ , and set $v_{1, ε} (t) : = \int_{0}^{t - ε} T (t - s) (f (s) - f (t)) d s$ . For $s \leq t - ε$ the semigroup law gives $T (t - s) = T (ε) T (t - s - ε)$ , so the integrand lies in $D (A)$ and, since $A T (ε)$ is bounded by [L1], Riemann sums and the norm inequality [L3] give $v_{1, ε} (t) \in D (A)$ and $A v_{1, ε} (t) = \int_{0}^{t - ε} A T (t - s) (f (s) - f (t)) d s$ .

1.2L2L3givenalgebra

The constant-endpoint term. Since $f (t)$ does not depend on the integration variable, $v_{2} (t) = \int_{0}^{t} T (τ) f (t) d τ = J_{t} f (t)$ , so [L2] gives $v_{2} (t) \in D (A)$ and $A v_{2} (t) = T (t) f (t) - f (t)$ , whence $∥ A v_{2} (t) ∥ \leq (c_{0} + 1) ∥ f ∥_{\infty}$ .

2.1step 1.1L1L3givenalgebra

The truncated first term is Cauchy in the graph norm. For $0 < ε < η < t$ the difference of the truncated $A$ -images is the integral over $[t - η, t - ε]$ of $A T (t - s) (f (s) - f (t))$ , whose norm is at most $c_{1} [f]_{α} (t - s)^{α - 1}$ by [L1] and Hölder continuity of $f$ ; integrating gives $∥ A v_{1, ε} (t) - A v_{1, η} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} (η^{α} - ε^{α}) \to 0$ as $ε, η ↓ 0$ . Likewise $∥ v_{1, ε} (t) - v_{1, η} (t) ∥ \leq c_{0} [f]_{α} \int_{t - η}^{t - ε} (t - s)^{α} d s \to 0$ , so both $v_{1, ε} (t)$ and $A v_{1, ε} (t)$ converge; since $A$ is closed, $v_{1} (t) \in D (A)$ and $A v_{1} (t) = \int_{0}^{t} A T (t - s) (f (s) - f (t)) d s$ , with $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ .

3.1step 1.2step 2.1L2L3givenalgebra

Continuity in the graph norm. The bounds just obtained give $∥ v_{1} (t) ∥ \leq c_{0} [f]_{α} t^{1 + α} / (1 + α)$ and $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ , so $v_{1}$ and $A v_{1}$ extend continuously to $t = 0$ with value $0$ ; on every $[a, b]$ with $a > 0$ , the truncated expressions are continuous for $0 < ε < a$ , and their tails are bounded uniformly in $t$ by $c_{0} [f]_{α} ε^{1 + α} / (1 + α)$ and $c_{1} [f]_{α} ε^{α} / α$ , respectively. They therefore converge uniformly on $[a, b]$ , proving continuity of $v_{1}$ and $A v_{1}$ at positive times. For the constant-endpoint term, $∥ A v_{2} (t) ∥ \leq (c_{0} + 1) ∥ f ∥_{\infty}$ and $∥ A v_{2} (t) - A v_{2} (0) ∥ \leq (c_{0} + 1) ∥ f (t) - f (0) ∥ + ∥ (T (t) - I) f (0) ∥ \to 0$ by continuity of $f$ and strong continuity of $T$ ; finally $v$ is continuous on $[0, b]$ by [L3]. Hence $v \in C ([0, b], D (A))$ in the graph norm and $A v \in C ([0, b], X)$ with $A v (0) = 0$ .

4.1step 1.2step 3.1L2givenalgebra∎

The final assertion. For $x \in D (A)$ , [L2] gives $A T (t) x = T (t) A x \to A x$ as $t ↓ 0$ by strong continuity, so $A u (t) = A T (t) x + A v (t) \to A x + 0 = A x$ ; the decomposition $u = T (\cdot) x + v$ therefore removes the singularity of $A T (t) x$ at the endpoint, and no choice principle beyond Dependent Choice was used.

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Compatibility at time zero for a classical parabolic solution

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 $A$ be the generator of a strongly continuous semigroup on a Banach space $X$ (Infinitesimal generator of a C0-semigroup, Banach space), let $b > 0$ , $x \in X$ , and let $f : [0, b] \to X$ be continuous with $f (0)$ defined. Suppose $u \in C ([0, b], X) \cap C^{1} ((0, b], X)$ satisfies $u (0) = x$ , $u (t) \in D (A)$ and $u^{'} (t) = A u (t) + f (t)$ for every $t \in (0, b]$ , and suppose $\lim_{t ↓ 0} u^{'} (t)$ exists in $X$ (in particular if $u \in C^{1} ([0, b], X)$ ). Then $x \in D (A)$ and $\lim_{t ↓ 0} u^{'} (t) = A x + f (0);$ consequently $u$ extends to a classical solution on $[0, b]$ in the sense of Classical, strong and mild abstract Cauchy solutions exactly when this limit exists, and in the PDE realisation $x \in D (A)$ is precisely the boundary-and-domain compatibility of the initial datum. One-order propagation under stronger regularity. If, in addition, $f \in C^{1} ([0, b], X)$ , $u \in C^{1} ((0, b], D (A))$ in the graph norm (Unbounded linear operators: domain, graph and extension), and $A x + f (0) \in D (A)$ with $u^{'} (t) \to A x + f (0)$ in the graph norm as $t ↓ 0$ , then $u^{'}$ is right-differentiable at $0$ with $(u^{'})^{'} (0) = A (A x + f (0)) + f^{'} (0)$ . No choice principle beyond Dependent Choice is used.

Facts & Assumptions

Given: A strongly continuous semigroup with generator $A$ on the Banach space $X$ , the closed operator $A$ with domain $D (A)$ , a continuous $f : [0, b] \to X$ with $f (0)$ defined, and $u \in C ([0, b], X) \cap C^{1} ((0, b], X)$ with $u (0) = x$ , $u (t) \in D (A)$ , $u^{'} (t) = A u (t) + f (t)$ on $(0, b]$ and $y : = \lim_{t ↓ 0} u^{'} (t)$ existing in $X$ . For the one-order propagation clause, also assume $f \in C^{1} ([0, b], X)$ , $u \in C^{1} ((0, b], D (A))$ in graph norm, $A x + f (0) \in D (A)$ , and $u^{'} (t) \to A x + f (0)$ in graph norm.

[L1]

$A$ is closed and densely defined: its graph ${(v, A v) : v \in D (A)}$ is closed in $X \times X$ (The generator is closed and densely defined).

[L2]

A classical solution on $[0, b]$ is a function in $C^{1} ([0, b], X)$ with values in $D (A)$ , $A u \in C ([0, b], X)$ , satisfying the equation on $(0, b)$ and the initial condition; continuous forcing on $[0, b]$ extends the equation to the endpoints (Classical, strong and mild abstract Cauchy solutions).

[L3]

Unbounded linear operators: domain, graph and extension: the graph norm on $D (A)$ is $∥ v ∥_{A} = (∥ v ∥^{2} + ∥ A v ∥^{2})^{1 / 2}$ , so $A : D (A)_{g r a p h} \to X$ is bounded.

[L4]

Fundamental theorem of calculus for Banach-valued continuous curves: if a continuous Banach-valued curve is differentiable on $(0, b)$ and its derivative extends continuously to $[0, b]$ , then its increment equals the integral of that derivative.

Proof

technique · direct

1.1givenalgebra

Limit of $A u (t)$ . For $t \in (0, b]$ the equation gives $A u (t) = u^{'} (t) - f (t)$ , and $u^{'} (t) \to y$ by the hypothesis while $f (t) \to f (0)$ by continuity; hence $A u (t) \to y - f (0)$ in $X$ .

1.2given

Limit of $u (t)$ . By continuity of $u$ at $0$ and $u (0) = x$ one has $u (t) \to x$ as $t ↓ 0$ (Fréchet derivative between Banach spaces), and $u (t) \in D (A)$ for every $t \in (0, b]$ .

2.1step 1.1step 1.2L1given

Closedness forces the endpoint compatibility. The pairs $(u (t), A u (t))$ lie in the graph of $A$ for $t \in (0, b]$ and converge to $(x, y - f (0))$ by [step 1.1] and [step 1.2]; since the graph is closed by [L1], the limit lies in the graph: $x \in D (A)$ and $A x = y - f (0)$ , that is $\lim_{t ↓ 0} u^{'} (t) = A x + f (0)$ .

3.1step 2.1L2L4given

Equivalence with the classical solution. If the limit exists, [step 2.1] shows $x \in D (A)$ and, using $A u (t) = u^{'} (t) - f (t) \to A x$ , the derivative $u^{'}$ extends continuously to $[0, b]$ with value $A x + f (0) = A u (0) + f (0)$ . By [L4] applied to $u$ on $[0, t]$ , $u (t) - u (0) = \int_{0}^{t} u^{'} (s) d s$ , so the right derivative at $0$ is this limiting value. Hence $u \in C^{1} ([0, b], X)$ solves the equation at every point of $[0, b]$ and is a classical solution by [L2]; conversely a classical solution has $u \in C^{1} ([0, b], X)$ , so its one-sided derivative at $0$ exists and the limit does.

3.2step 2.1L3L4givenalgebra

(One-order propagation under stronger regularity) Assume the additional hypotheses in the final Statement clause and put $w (0) : = A x + f (0)$ , $w (t) : = u^{'} (t)$ for $t > 0$ . By [L3], $A$ is bounded from the graph-norm domain into $X$ ; since $u$ is $C^{1}$ there in graph norm and $f \in C^{1}$ , differentiating $u^{'} = A u + f$ on $(0, b]$ gives $w^{'} (t) = A w (t) + f^{'} (t)$ . The graph-norm convergence of $w$ and continuity of $f^{'}$ imply $w^{'} (t) \to A (A x + f (0)) + f^{'} (0)$ . By [L4] on $[0, t]$ , $w (t) - w (0) = \int_{0}^{t} w^{'} (s) d s$ ; dividing by $t$ and using continuity of the integrand at $0$ gives the stated right derivative of $u^{'}$ at $0$ .

4.1step 2.1step 3.1step 3.2given∎

Assembly. [step 2.1] proves $x \in D (A)$ and the value of the limit; [step 3.1] gives the stated equivalence with the classical solution; [step 3.2] proves the one-order propagation. The argument used only continuity, closedness of the graph and the equation, so no choice principle beyond Dependent Choice was used.

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Classical regularity for Holder-continuous forcing under initial compatibility

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 $A$ be sectorial of angle $δ \in (0, π / 2]$ in the $e^{t A}$ convention on a complex Banach space $X$ , with generated analytic semigroup $(T (t))_{t \geq 0}$ (Sectorial operator with the semigroup sign convention, Smoothing estimates for the semigroup generated by a sectorial operator). Let $b > 0$ , $x \in D (A)$ , and $f \in C^{α} ([0, b], X)$ for some $α \in (0, 1)$ . Define $u (t) : = T (t) x + \int_{0}^{t} T (t - s) f (s) d s, 0 \leq t \leq b .$ Then $u$ is a classical solution of $u^{'} = A u + f$ on $[0, b]$ in the sense of Classical, strong and mild abstract Cauchy solutions: $u \in C^{1} ([0, b], X)$ , $u (t) \in D (A)$ for every $t \in [0, b]$ , $u (0) = x$ , and $u^{'} (t) = A u (t) + f (t)$ for every $t \in [0, b]$ . Moreover $A u \in C ([0, b], X)$ and $A u (0) = A x$ . Set $c_{0} : = \sup_{0 \leq t \leq b} ∥ T (t) ∥$ and $c_{1} : = \sup_{0 < t \leq b} t ∥ A T (t) ∥$ . For every $0 < t \leq b$ , $A u (t) - A x = (T (t) - I) (A x + f (0)) + R_{f} (t),$ where $R_{f} (t) : = \int_{0}^{t} A T (t - s) (f (s) - f (t)) d s + (T (t) - I) (f (t) - f (0)),$ and $∥ R_{f} (t) ∥ \leq (\frac{c_{1}}{α} + c_{0} + 1) [f]_{α} t^{α}$ . Thus the endpoint modulus includes the semigroup orbit of $A x + f (0)$ ; the stated hypotheses alone give no Hölder modulus for $A u$ in terms of $[f]_{α}$ alone.

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ$ with its analytic semigroup $T$ on the complex Banach space $X$ , constants $c_{0}, c_{1}$ as above, $b > 0$ , $x \in D (A)$ , $f \in C^{α} ([0, b], X)$ with Hölder constant $[f]_{α}$ and $α \in (0, 1)$ , and $u (t) : = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ with $v (t) : = u (t) - T (t) x$ .

[L1]

$v \in C ([0, b], D (A))$ in the graph norm, $A v \in C ([0, b], X)$ with $v (0) = 0$ and $A v (0) = 0$ , and $A v (t) = A v_{1} (t) + A v_{2} (t)$ with $A v_{1} (t) = \int_{0}^{t} A T (t - s) (f (s) - f (t)) d s$ , $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ and $A v_{2} (t) = (T (t) - I) f (t)$ (Analytic Duhamel cancellation removes the generator singularity).

[L2]

The variation-of-constants formula makes $u$ the unique integral solution of $u^{'} = A u + f$ , and the integral-solution identity together with $A \int_{0}^{t} T (s) x d s = T (t) x - x$ gives $v (t) = A \int_{0}^{t} v (s) d s + \int_{0}^{t} f (s) d s$ for $t \in [0, b]$ (Variation of constants for the inhomogeneous abstract Cauchy problem, Classical, strong and mild abstract Cauchy solutions, Time integrals of semigroup orbits lie in the generator domain); moreover $A T (t) y = T (t) A y$ for $y \in D (A)$ and $T (t) X \subseteq D (A)$ for $t > 0$ (The generator commutes with the semigroup on its domain, Smoothing estimates for the semigroup generated by a sectorial operator).

[L3]

For a continuous curve $g : [0, b] \to X$ the primitive $G (t) = \int_{0}^{t} g$ is differentiable with $G^{'} = g$ and is $C^{1}$ when $g$ is continuous (Fundamental theorem of calculus for Banach-valued continuous curves).

[L4]

For a sectorial operator $B$ with vertex $0$ , the contour semigroup $S$ is bounded on positive real times, has generator $B$ , is unique among exponentially bounded semigroups with that generator, and satisfies $S (t) X \subseteq D (B)$ and $∥ B S (t) ∥ \leq K_{1} / t$ (The generator of the contour semigroup is the sectorial operator, Smoothing estimates for the semigroup generated by a sectorial operator). Every strongly continuous semigroup has an exponential bound under DC (Exponential bound for a C0-semigroup).

Proof

technique · direct

1.1L4givenalgebra

Finite-interval smoothing. Choose a sectorial vertex $ω$ for $A$ and set $B = A - ω I$ on $D (A)$ . The identity $R (λ, B) = R (λ + ω, A)$ makes $B$ sectorial with vertex $0$ . The strongly continuous semigroup $S (t) = e^{- ω t} T (t)$ has generator $B$ on exactly $D (A)$ , because $(S (h) y - y) / h = e^{- ω h} (T (h) y - y) / h + (e^{- ω h} - 1) y / h$ . By [L4] it is exponentially bounded and equals the contour semigroup of $B$ . Writing $K_{0} = \sup_{t \geq 0} ∥ S (t) ∥$ and using $A T (t) = e^{ω t} (B S (t) + ω S (t))$ gives $T (t) X \subseteq D (A)$ for $t > 0$ , $c_{0} \leq e^{\max {ω, 0} b} K_{0}$ and $c_{1} \leq e^{\max {ω, 0} b} (K_{1} + ∣ ω ∣ b K_{0}) < \infty$ . These finite-interval bounds meet the Duhamel cancellation hypotheses and supply the positive-time domain inclusion for the given vertex.

2.1L1L2step 1.1given

The Duhamel data. By [L1] the function $v$ is continuous in the graph norm and $A v$ is continuous on $[0, b]$ with $A v (0) = 0$ , so $s \mapsto A v (s) + f (s)$ is a continuous $X$ -valued curve; moreover by [L1] the decomposition $A v = A v_{1} + A v_{2}$ holds with $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ and $∥ A v_{2} (t) ∥ \leq (c_{0} + 1) ∥ f ∥_{\infty}$ , and $∥ (T (t) - I) (f (t) - f (0)) ∥ \leq (c_{0} + 1) [f]_{α} t^{α}$ .

3.1step 2.1L1L2L3givenalgebra

The integral identity. By [L2] one has $v (t) = A \int_{0}^{t} v (s) d s + \int_{0}^{t} f (s) d s$ ; since $v$ and $A v$ are continuous on $[0, b]$ , the graph-norm integral $\int_{0}^{t} v (s) d s$ lies in $D (A)$ with $A \int_{0}^{t} v (s) d s = \int_{0}^{t} A v (s) d s$ , because $A$ is closed and the Riemann sums of the $D (A)$ -valued continuous curve $s \mapsto v (s)$ converge in the graph norm. Hence $v (t) = \int_{0}^{t} (A v (s) + f (s)) d s$ with a continuous integrand.

4.1step 2.1step 3.1L1L2L3givenalgebra

Classicality. The fundamental theorem [L3] applied to the continuous curve $s \mapsto A v (s) + f (s)$ shows $v \in C^{1} ([0, b], X)$ with $v^{'} = A v + f$ and $v (0) = 0$ . For $t > 0$ the orbit $T (t) x$ has derivative $A T (t) x = T (t) A x$ by [L2] and this derivative extends continuously to $0$ with value $A x$ because $x \in D (A)$ and $T$ is strongly continuous; hence $u = T (\cdot) x + v \in C^{1} ([0, b], X)$ with $u (0) = x$ and $u^{'} (t) = A T (t) x + A v (t) + f (t) = A u (t) + f (t)$ for every $t \in [0, b]$ , and $u (t) \in D (A)$ because both $T (t) x$ and $v (t)$ lie in the domain. Thus $u$ is a classical solution in the sense of the cited definition, and $A u \in C ([0, b], X)$ with $A u (0) = A x$ .

5.1step 2.1step 4.1L1L2given∎

Endpoint identity, modulus and caveat. Subtracting $A x$ from $A u (t) = A T (t) x + A v_{1} (t) + A v_{2} (t)$ and writing $A v_{2} (t) = (T (t) - I) f (t) = (T (t) - I) f (0) + (T (t) - I) (f (t) - f (0))$ gives $A u (t) - A x = (T (t) - I) (A x + f (0)) + R_{f} (t)$ with $R_{f}$ as displayed, and the bounds of [step 2.1] give $∥ R_{f} (t) ∥ \leq (\frac{c_{1}}{α} + c_{0} + 1) [f]_{α} t^{α}$ . The term $(T (t) - I) (A x + f (0))$ tends to $0$ by strong continuity but admits no uniform power modulus as stated, so the hypotheses give continuity and classicality of $u$ but not Hölder continuity of $A u$ in terms of $[f]_{α}$ alone. The argument used only the Duhamel cancellation, the variation-of-constants identity and the fundamental theorem, so no choice principle beyond Dependent Choice was used.

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Abstract parabolic smoothing for mild solutions

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 $A$ be sectorial of angle $δ \in (0, π / 2]$ on a complex Banach space $X$ with generated analytic semigroup $(T (t))_{t \geq 0}$ (Sectorial operator with the semigroup sign convention). For $k \geq 0$ , set $D (A^{0}) : = X$ , define $D (A^{k}) : = {y \in D (A^{k - 1}) : A^{k - 1} y \in D (A)}$ for $k \geq 1$ , and give $D (A^{k})$ the graph norm $∥ y ∥_{D (A^{k})} : = \sum_{j = 0}^{k} ∥ A^{j} y ∥$ . Let $m \geq 1$ , $b > 0$ , $x \in X$ , and $f \in C^{m - 1, α} ([0, b], D (A^{m - 1}))$ for some $α \in (0, 1)$ , with time regularity measured in that graph norm. For $m = 1$ , impose no compatibility condition on $x$ . For $m \geq 2$ , define $γ_{0} : = x$ and $γ_{j + 1} : = A γ_{j} + f^{(j)} (0)$ for $0 \leq j < m - 1$ , and assume $γ_{j} \in D (A)$ for $0 \leq j \leq m - 1$ . Let $u (t) : = T (t) x + \int_{0}^{t} T (t - s) f (s) d s .$ Then $u (t) \in D (A^{m})$ for every $t \in (0, b]$ , $A^{m} u \in C ((0, b], X)$ , and, writing $g (t) : = A^{m - 1} f (t)$ , $∥ A^{m} u (t) ∥ \leq C_{m} t^{- m} ∥ x ∥ + (c_{0} + 1) ∥ g (0) ∥ + (\frac{c_{1}}{α} + c_{0} + 1) [g]_{α} t^{α}, 0 < t \leq b,$ where $c_{0} = \sup_{[0, b]} ∥ T (t) ∥$ , $c_{1} = \sup_{0 < t \leq b} t ∥ A T (t) ∥$ , and $C_{m}$ is the analytic smoothing constant for $A^{m} T (t)$ . In particular, for a constant $C_{m}^{'}$ depending only on $m, α, b$ and the semigroup bounds, $∥ A^{m} u (t) ∥ \leq C_{m}^{'} t^{- m} (∥ x ∥ + ∥ f ∥_{C^{m - 1, α} ([0, b], D (A^{m - 1}))}) .$ The compatibility tower is retained from the planned statement for $m \geq 2$ ; under this stronger graph-norm source hypothesis the proof below does not need the tower. For $m = 1$ , homogeneous smoothing gives the result for every $x \in X$ .

Facts & Assumptions

Given: A sectorial operator $A$ of angle $δ \in (0, π / 2]$ on the complex Banach space $X$ with generated analytic semigroup $T$ and constants $c_{0} = \sup_{0 \leq t \leq b} ∥ T (t) ∥$ , $c_{1} = \sup_{0 < t \leq b} t ∥ A T (t) ∥$ ; the recursively defined graph domains $D (A^{k})$ with $D (A^{0}) = X$ and $D (A^{k}) = {y \in D (A^{k - 1}) : A^{k - 1} y \in D (A)}$ and norms $∥ y ∥_{D (A^{k})} = \sum_{j = 0}^{k} ∥ A^{j} y ∥$ ; $m \geq 1$ , $b > 0$ , $x \in X$ , $α \in (0, 1)$ , $f \in C^{m - 1, α} ([0, b], D (A^{m - 1}))$ , $g : = A^{m - 1} f$ , and $u (t) : = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ ; for $m \geq 2$ the tower $γ_{0} = x$ , $γ_{j + 1} = A γ_{j} + f^{(j)} (0)$ is defined with $γ_{j} \in D (A)$ for $0 \leq j \leq m - 1$ (an unused hypothesis).

[L1]

For a sectorial operator $B$ with vertex $0$ , its contour semigroup $S$ satisfies $S (t) X \subseteq D (B^{k})$ , $∥ B^{k} S (t) ∥ \leq K_{k} t^{- k}$ for $k \geq 1$ , and $S^{(k)} (t) = B^{k} S (t)$ in operator norm (Smoothing estimates for the semigroup generated by a sectorial operator). It is bounded on the positive real axis, has generator $B$ , and is unique among exponentially bounded semigroups with that generator (The generator of the contour semigroup is the sectorial operator). Every strongly continuous semigroup has an exponential bound under the assumed Dependent Choice (Exponential bound for a C0-semigroup).

[L2]

For every $y \in D (A)$ and every $t \geq 0$ one has $A T (t) y = T (t) A y$ (The generator commutes with the semigroup on its domain).

[L3]

The generated semigroup $T$ is strongly continuous on $[0, \infty)$ with generator $A$ , and $A$ is closed (Complex sector and bounded analytic semigroup, Sectorial operator with the semigroup sign convention).

[L4]

If $x_{0} \in D (A)$ and $h \in C^{α} ([0, b], X)$ , then $U (t) : = T (t) x_{0} + \int_{0}^{t} T (t - s) h (s) d s$ is a classical solution: $U \in C^{1} ([0, b], X)$ , $U (t) \in D (A)$ for every $t$ , $U (0) = x_{0}$ , $U^{'} = A U + h$ pointwise, and $A U \in C ([0, b], X)$ with $A U (0) = A x_{0}$ (Classical regularity for Holder-continuous forcing under initial compatibility).

[L5]

For that $U$ , every $0 < t \leq b$ satisfies $A U (t) - A x_{0} = (T (t) - I) (A x_{0} + h (0)) + R_{h} (t)$ where $∥ R_{h} (t) ∥ \leq (\frac{c_{1}}{α} + c_{0} + 1) [h]_{α} t^{α}$ , with $c_{0}, c_{1}$ the semigroup constants and $[h]_{α}$ the Hölder constant of $h$ (Classical regularity for Holder-continuous forcing under initial compatibility).

Proof

technique · direct

1.1L1givenalgebra

Homogeneous term with an arbitrary vertex. Let $ω$ be a sectorial vertex for $A$ and put $B : = A - ω I$ . Since $R (λ, B) = R (λ + ω, A)$ , $B$ is sectorial with vertex $0$ . The semigroup $S (t) : = e^{- ω t} T (t)$ has generator $B$ on $D (A)$ : its difference quotient converges exactly when that of $T$ does, since $(S (h) y - y) / h = e^{- ω h} (T (h) y - y) / h + (e^{- ω h} - 1) y / h$ . It is exponentially bounded by [L1], hence equals the contour semigroup of $B$ by [L1]. Induction using $B = A - ω I$ gives $D (B^{k}) = D (A^{k})$ and $A^{k} = (B + ω I)^{k} = \sum_{j = 0}^{k} (\binom{k}{j}) ω^{k - j} B^{j}$ on this common domain: in the induction step, the lower powers $B^{j} y$ for $j < k$ already lie in $D (B) = D (A)$ , so $A^{k} y \in D (A)$ is equivalent to $B^{k} y \in D (B)$ . Put $K_{0} : = \sup_{t \geq 0} ∥ S (t) ∥$ . For $0 < t \leq b$ , [L1] now gives $T (t) X \subseteq D (A^{m})$ and $∥ A^{m} T (t) ∥ \leq e^{\max {ω, 0} b} \sum_{j = 0}^{m} (\binom{m}{j}) ∣ ω ∣^{m - j} K_{j} b^{m - j} t^{- m} = : C_{m} t^{- m}$ . In particular $c_{0}, c_{1}$ are finite. Differentiating $T (t) = e^{ω t} S (t)$ gives $T^{(m)} (t) = A^{m} T (t)$ , continuous in operator norm for $t > 0$ . This $C_{m}$ is a finite-interval smoothing constant; no global $t^{- m}$ bound for a nonzero vertex is asserted. Writing $u = T (\cdot) x + v$ , with $v (t) : = \int_{0}^{t} T (t - s) f (s) d s$ , reduces the remaining membership, continuity and estimate to those for $A^{m} v$ .

1.2L2L3givenalgebra

The curves $h_{j}$ and their images. The graph norm on $D (A^{m - 1})$ dominates $∥ \cdot ∥$ and $∥ A^{j} \cdot ∥$ for every $0 \leq j \leq m - 1$ , so each $h_{j} : = A^{j} f$ is a continuous $X$ -valued curve on $[0, b]$ ; the curve $g = h_{m - 1}$ satisfies $∥ g (0) ∥ \leq ∥ f (0) ∥_{D (A^{m - 1})}$ and, for $m \geq 2$ , is differentiable in $X$ with $g^{'} = A^{m - 1} f^{'}$ bounded, hence Lipschitz and α-Hölder, while for $m = 1$ it equals $f$ and is α-Hölder by hypothesis; thus $g \in C^{α} ([0, b], X)$ with $∥ g (0) ∥$ and $[g]_{α}$ controlled by $∥ f ∥_{C^{m - 1, α} ([0, b], D (A^{m - 1}))}$ . Fix $0 < t \leq b$ and $0 \leq j \leq m - 2$ and put $u_{j} (s) : = T (t - s) h_{j} (s)$ : by strong continuity of $T$ from [L3] and continuity of $h_{j}$ the curve $u_{j}$ is continuous on $[0, t]$ , and for every $s \in [0, t]$ one has $h_{j} (s) \in D (A)$ , $u_{j} (s) \in D (A)$ and $A u_{j} (s) = T (t - s) h_{j + 1} (s) = u_{j + 1} (s)$ by [L2].

2.1step 1.2L3L4givenalgebra

Domain induction by closedness. The claim is that for every $0 \leq j \leq m - 1$ one has $v (t) \in D (A^{j})$ with $A^{j} v (t) = V_{j} (t) : = \int_{0}^{t} T (t - s) h_{j} (s) d s$ ; the case $j = 0$ is the definition of $v$ . Assume the claim for some $j \leq m - 2$ and take right-endpoint Riemann sums $S_{n}$ of the continuous curve $u_{j}$ along partitions of $[0, t]$ with mesh tending to $0$ : then $S_{n} \to V_{j} (t) = A^{j} v (t)$ , while each $S_{n}$ lies in $D (A)$ and $A S_{n}$ is the corresponding Riemann sum of $u_{j + 1}$ , so $A S_{n} \to V_{j + 1} (t)$ by [step 1.2]; since $A$ is closed by [L3], $V_{j} (t) \in D (A)$ and $A V_{j} (t) = V_{j + 1} (t)$ , that is $v (t) \in D (A^{j + 1})$ and $A^{j + 1} v (t) = V_{j + 1} (t)$ . Induction up to $j = m - 1$ gives $v (t) \in D (A^{m - 1})$ and $A^{m - 1} v (t) = w (t) : = \int_{0}^{t} T (t - s) g (s) d s$ , which is exactly the function $U$ of [L4] with $x_{0} = 0$ and $h = g$ .

3.1step 1.2step 2.1L4L5givenalgebra

Classical regularity of $w$ . By [step 1.2] the forcing $g$ lies in $C^{α} ([0, b], X)$ , so [L4] applied to $x_{0} = 0$ and $h = g$ makes $w$ a classical solution with $w (t) \in D (A)$ for every $t \in [0, b]$ , $A w \in C ([0, b], X)$ and $A w (0) = 0$ , and [L5] gives $A w (t) = (T (t) - I) g (0) + R_{g} (t)$ with $∥ R_{g} (t) ∥ \leq (\frac{c_{1}}{α} + c_{0} + 1) [g]_{α} t^{α}$ for $0 < t \leq b$ ; since $A^{m - 1} v (t) = w (t)$ by [step 2.1], the recursive definition of $D (A^{m})$ yields $v (t) \in D (A^{m})$ with $A^{m} v (t) = A w (t)$ .

4.1step 1.1step 1.2step 3.1givenalgebra∎

Final estimate and continuity. Adding the homogeneous bound of [step 1.1] to the bound of [step 3.1] gives, for every $0 < t \leq b$ , $∥ A^{m} u (t) ∥ \leq C_{m} t^{- m} ∥ x ∥ + (c_{0} + 1) ∥ g (0) ∥ + (\frac{c_{1}}{α} + c_{0} + 1) [g]_{α} t^{α}$ , and $A^{m} u$ is continuous on $(0, b]$ because both $A^{m} T (\cdot) x$ and $A w$ are continuous there by [step 1.1] and [step 3.1]; since $t \leq b$ and the norms of $g (0)$ and $[g]_{α}$ are controlled by [step 1.2], this gives $∥ A^{m} u (t) ∥ \leq C_{m}^{'} t^{- m} (∥ x ∥ + ∥ f ∥_{C^{m - 1, α} ([0, b], D (A^{m - 1}))})$ with $C_{m}^{'}$ depending only on $m, α, b$ and the semigroup bounds. For $m = 1$ the same argument runs with the single curve $h_{0} = f$ and the vacuous tower, and the splitting of [step 1.1] is what removes every requirement on $x \in X$ ; no choice principle beyond Dependent Choice is used.

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Abstract generator-domain smoothing becomes spatial regularity only after domain identification

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.

The homogeneous smoothing theorem Smoothing estimates for the semigroup generated by a sectorial operator says that for every $x \in X$ and $t > 0$ , $T (t) x \in D (A^{m})$ for each $m \geq 1$ . For a forced mild solution $u (t) = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ , the positive-time $D (A^{m})$ conclusion of Abstract parabolic smoothing for mild solutions uses the stated source hypothesis $f \in C^{m - 1, α} ([0, b], D (A^{m - 1}))$ in the graph norm; it is not asserted for arbitrary $X$ -valued forcing. The graph-domain membership is a statement about an abstract operator on a Banach space; it names a Sobolev derivative only after an elliptic-regularity theorem identifies $D (A^{m})$ with a concrete space. For the Dirichlet Laplacian on a bounded $C^{2}$ domain one has $D (A) = H^{2} \cap H_{0}^{1}$ by Global $H^{2}$ Dirichlet regularity, and for a $C^{2 m}$ boundary $D (A^{m}) = {u \in H^{2 m} : Δ^{j} u \in H_{0}^{1}, 0 \leq j < m}$ by Higher-order boundary regularity for Dirichlet problems; without the boundary compatibility hypotheses, only the recursive graph domain is available (the identification clauses of The Dirichlet Laplacian generates an analytic heat semigroup). Likewise, a diagonal analytic semigroup on a sequence space smooths homogeneous orbits into powers of a sequence operator with no intrinsic spatial variables (the abstract sequence-space example of the companion page). This remark is not proof-bearing; it fixes the seam between the abstract theory and its PDE realisations.

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Real Banach spaces require complexification for analyticity

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.

Given a real Banach space $X$ and a strongly continuous semigroup $(T (t))_{t \geq 0}$ of bounded real-linear operators, first complexify each real-time operator to $T_{C} (t) : = (T (t))_{C}$ on $X_{C} = X \times X$ with the rotation-supremum norm (Canonical Banach complexification of a real Banach space). The real semigroup is analytic of angle $δ$ when this complexified real-time semigroup admits an analytic extension ${\tilde{T}}_{C} (z)$ to $Σ_{δ} \cup {0}$ on $X_{C}$ , agreeing with $T_{C} (t)$ for $t \geq 0$ ; the real-time operators preserve the embedded copy $X \times {0}$ ; preservation at nonreal times is not guaranteed, but can occur (the identity semigroup preserves it at every complex time) (Complex sector and bounded analytic semigroup). It is bounded analytic when the extension is bounded on every strictly smaller sector. A holomorphic map is complex-time: a real-linear family defined only for real $t$ is not itself a map on a complex sector.

For a bounded real-linear operator its spectrum and resolvent are computed on its complexification using Resolvent and spectrum of a closed operator on a Banach space; no spectral-radius assertion is used here. For an unbounded real generator $A$ , complexify its domain and action, $A_{C} (x, y) = (A x, A y)$ on $D (A) \times D (A) \subset X_{C}$ , then impose the closed-unbounded resolvent and sectorial conditions on $A_{C}$ using Resolvent and spectrum of a closed operator on a Banach space and Sectorial operator with the semigroup sign convention. The sectorial-generation theorem Sectorial resolvent characterisation of bounded analytic semigroups is applied to that complexified operator. The heat equation on real $L^{2}$ is recovered by restricting $T_{C} (t)$ to the real summand for real $t \geq 0$ ; its holomorphic extension is on the complexification, not a real-valued map at nonreal times.

5 · Examples, counterexamples and false statements

None yet.