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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]

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).

[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.

Depends on

Used by

Form-generated sectorial elliptic semigroups Theorem

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Sources

Roland Schnaubelt, Evolution Equations, Karlsruhe Institute of Technology (2023/24 course, complete lecture notes) (standard reference, not scraped)
Klaus-Jochen Engel and Rainer Nagel, One-Parameter Semigroups for Linear Evolution Equations, Graduate Texts in Mathematics 194 (complete author-hosted monograph) (standard reference, not scraped)