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Lax--Milgram and Weak Elliptic Solutions

1 · Prerequisites

2 · Summary

This page builds the abstract machinery of the Lax--Milgram theorem and applies it to weak elliptic Dirichlet problems. Bounded, coercive and symmetric sesquilinear forms and the negative Sobolev space $H^{- 1} (Ω)$ are defined first; a bounded form is represented by a unique bounded operator $A$ with $a (u, v) = (A u, v)$ , and the adjoint of a coercive form is coercive with the same constants. Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence, and conversely $L^{2}$ forcing embeds boundedly in $H^{- 1}$ with the Poincar'e constant. Coercivity makes $A$ bounded below, and a small form step $I - ρ A$ is a strict contraction; the Banach fixed-point theorem then proves Lax--Milgram, with solution norm at most $∥ F ∥ / α$ , and an energy expansion characterises symmetric solutions as energy minimisers. For uniformly elliptic divergence-form operators the form is bounded and coercive on $H_{0}^{1}$ under the explicit smallness condition $θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0$ , giving unique weak solutions of $L u = F$ and of the zero-boundary Poisson problem; a positive reaction term supplies a second coercivity mechanism on all of $H^{1}$ , while on a nonempty bounded connected $W^{1, 2}$ -extension domain the Neumann problem is solved on the mean-zero subspace exactly under the compatibility condition $F (1) = 0$ . Classical solutions are weak solutions, inhomogeneous boundary data are handled by a trace lifting, and weak solutions depend Lipschitz-continuously on their data.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Bounded, coercive and symmetric sesquilinear forms

Definition

Let $H$ be a real or complex Hilbert space over $K \in {R, C}$ with inner product $(\cdot, \cdot)$ linear in the first argument and conjugate-linear in the second (Real and complex inner-product spaces and their induced length, Hilbert space), and let $a : H \times H \to K$ be sesquilinear in the sense of Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable: linear in the first argument and conjugate-linear in the second. The form $a$ is bounded with bound $M \geq 0$ when $∣ a (u, v) ∣ \leq M ∥ u ∥ ∥ v ∥ for all u, v \in H,$ and coercive with constant $α > 0$ when $Re a (u, u) \geq α ∥ u ∥^{2} for all u \in H .$ It is symmetric (Hermitian) when $a (u, v) = \overline{a (v, u)}$ for all $u, v$ , over $C$ this is equivalent to $a (u, u) \in R$ for every $u$ . Indeed, writing $q (u) = a (u, u)$ , sesquilinearity gives $4 a (u, v) = q (u + v) - q (u - v) + i q (u + i v) - i q (u - i v)$ , and real diagonal values make this identity conjugate-symmetric. Over $R$ , symmetry means $a (u, v) = a (v, u)$ ; real diagonal values alone do not imply symmetry. The adjoint form is $a^{*} (u, v) : = \overline{a (v, u)}$ , and $(a^{*})^{*} = a$ . Real bilinear convention. When $H$ is a real Hilbert space the same definitions apply with $a$ bilinear and coercive in the form $a (u, u) \geq α ∥ u ∥^{2}$ ; the conjugation in the second slot is then the identity. More generally, a real bilinear form may satisfy the boundedness and coercivity conditions without being symmetric; symmetry is an additional property, not part of either condition. When $H = {0}$ every form is bounded with bound $0$ and coercive with every $α > 0$ ; this degenerate case is kept but is never load-bearing. All constants below are named and never silently improved. No choice principle is used in this definition.

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

A bounded form is represented by a unique bounded operator

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), used through Riesz representation for Hilbert spaces. Let $H$ be a real or complex Hilbert space and let $a$ be a bounded sesquilinear form on $H$ with bound $M$ in the sense of Bounded, coercive and symmetric sesquilinear forms. Then there is a unique bounded linear operator $A \in B (H)$ with $a (u, v) = (A u, v) for all u, v \in H,$ and $∥ A ∥ \leq M$ ; if $M$ is the least bound of $a$ then $∥ A ∥ = M$ . The map $a \mapsto A$ is linear, and $a$ is coercive with constant $α$ if and only if $Re (A u, u) \geq α ∥ u ∥^{2}$ for all $u$ . For the adjoint form one has $a^{*} (u, v) = (A^{*} u, v)$ , where $A^{*}$ is the Hilbert adjoint of The Hilbert-space adjoint of a bounded operator.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ with inner product $(\cdot, \cdot)$ linear in the first argument and conjugate-linear in the second; a sesquilinear form $a$ on $H$ , linear in the first argument and conjugate-linear in the second, with bound $M \geq 0$ .

[F1]

Bounded and sesquilinear: $a (u, λ v) = \overline{λ} a (u, v)$ , $a (λ u, v) = λ a (u, v)$ , $a (u + u^{'}, v) = a (u, v) + a (u^{'}, v)$ , and $∣ a (u, v) ∣ \leq M ∥ u ∥ ∥ v ∥$ for all $u, v \in H$ (Bounded, coercive and symmetric sesquilinear forms).

[F2]

Inner-product facts: $(v, w) = \overline{(w, v)}$ , positive definiteness (so a vector orthogonal to all of $H$ is $0$ ), and Cauchy--Schwarz $∣ (u, v) ∣ \leq ∥ u ∥ ∥ v ∥$ ; the inner product is linear in the first slot and conjugate-linear in the second (Real and complex inner-product spaces and their induced length, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Hilbert space).

[F3]

Riesz representation under Countable Choice: every bounded linear functional $f$ on $H$ has a unique $y \in H$ with $f (x) = (x, y)$ for all $x$ , and $∥ f ∥ = ∥ y ∥$ (Riesz representation for Hilbert spaces, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

Bounded operators and the operator norm: $T$ is bounded when some $C \geq 0$ has $∥ T x ∥ \leq C ∥ x ∥$ , and $∥ T ∥ = \sup_{∥ x ∥ \leq 1} ∥ T x ∥$ is its least bound (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F5]

Hilbert adjoint: there is a unique $A^{*} \in B (H)$ with $(A u, v) = (u, A^{*} v)$ for all $u, v \in H$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

Proof

1.1F1F2

For fixed $u \in H$ the map $f_{u} (v) : = \overline{a (u, v)}$ is linear in $v$ and bounded: $f_{u} (λ v) = \overline{a (u, λ v)} = \overline{\overline{λ} a (u, v)} = λ f_{u} (v)$ and, more generally, conjugate-linearity of $a$ in the second slot makes $f_{u}$ additive, while $∣ f_{u} (v) ∣ = ∣ a (u, v) ∣ \leq M ∥ u ∥ ∥ v ∥$ shows that $∥ f_{u} ∥ \leq M ∥ u ∥$ .

2.1F2F3step 1.1

Riesz representation defines $A$ : by [F3] there is a unique $A u \in H$ with $f_{u} (v) = (v, A u)$ for every $v$ , that is $\overline{a (u, v)} = (v, A u)$ , and $∥ A u ∥ = ∥ f_{u} ∥ \leq M ∥ u ∥$ . Conjugating the representing identity with the conjugate symmetry of the inner product gives $a (u, v) = \overline{(v, A u)} = (A u, v)$ for all $v$ ; so every $u$ is assigned a unique vector $A u$ with $a (u, v) = (A u, v)$ for all $u, v$ , and in particular $∥ A u ∥ \leq M ∥ u ∥$ .

3.1F1F2F4step 2.1

$A$ is linear: for scalars $s, t$ and $u, w \in H$ , first-slot linearity of $a$ gives $a (s u + t w, v) = s a (u, v) + t a (w, v)$ for every $v$ , hence $(A (s u + t w), v) = s (A u, v) + t (A w, v) = (s A u + t A w, v)$ by linearity of the inner product in its first slot, and positive definiteness forces $A (s u + t w) = s A u + t A w$ . Therefore $A$ is linear and, by step 2.1, bounded with $∥ A ∥ \leq M$ .

3.2F2step 2.1

$A$ is unique: if $B \in B (H)$ also satisfies $a (u, v) = (B u, v)$ for all $u, v$ , then $(A u - B u, v) = 0$ for every $v$ , and positive definiteness gives $A u = B u$ for every $u$ , that is $A = B$ . The assignment $a \mapsto A$ is linear: for forms $a, b$ with operators $A, B$ and a scalar $c$ , $(A_{a + c b} u, v) = (a + c b) (u, v) = (A u, v) + c (B u, v) = ((A + c B) u, v)$ for all $v$ , so $A_{a + c b} = A + c B$ by the same uniqueness argument.

4.1F1F2F4step 3.1algebra

Least bound and coercivity: if $M$ is the least bound of $a$ , then for all $u, v \in H$ one has $∣ a (u, v) ∣ = ∣ (A u, v) ∣ \leq ∥ A u ∥ ∥ v ∥ \leq ∥ A ∥ ∥ u ∥ ∥ v ∥$ , so $∥ A ∥$ is itself a bound of $a$ and $M \leq ∥ A ∥$ ; with step 3.1 this gives $∥ A ∥ = M$ . Also, substituting $v = u$ in $a (u, v) = (A u, v)$ gives $a$ coercive with constant $α > 0$ if and only if $Re (A u, u) \geq α ∥ u ∥^{2}$ for every $u$ .

5.1F2F5∎

Adjoint form: for all $u, v \in H$ , $a^{*} (u, v) = \overline{a (v, u)} = \overline{(A v, u)} = (u, A v) = (A^{*} u, v)$ by conjugate symmetry of the inner product and the defining identity $(A v, u) = (v, A^{*} u)$ of the Hilbert adjoint; hence the adjoint form is represented by $A^{*}$ .

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

A coercive form operator is bounded below

Statement

Assume Countable Choice, used through A bounded form is represented by a unique bounded operator. Let $a$ be a bounded coercive sesquilinear form on a real or complex Hilbert space $H$ with constants $M, α$ (Bounded, coercive and symmetric sesquilinear forms) and let $A$ be its operator, $a (u, v) = (A u, v)$ . Then $α ∥ u ∥ \leq ∥ A u ∥ \leq M ∥ u ∥ for every u \in H,$ so $A$ is injective and bounded below with constant $α$ in the sense of A bounded operator that is bounded below.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ on $H$ with bound $M \geq 0$ and coercivity constant $α > 0$ ; and its operator $A \in B (H)$ , $a (u, v) = (A u, v)$ .

[F1]

$A$ exists, is linear and bounded with $a (u, v) = (A u, v)$ for all $u, v$ and $∥ A ∥ \leq M$ ; $a$ is linear in the first argument and conjugate-linear in the second, and $Re a (u, u) \geq α ∥ u ∥^{2}$ (Bounded, coercive and symmetric sesquilinear forms, A bounded form is represented by a unique bounded operator).

[F2]

Cauchy--Schwarz: $∣ (x, y) ∣ \leq ∥ x ∥ ∥ y ∥$ ; and for a complex number $z$ one has $Re z \leq ∣ z ∣$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Real and imaginary parts, complex conjugation, and modulus).

[F3]

$T \in B (X, Y)$ is bounded below when $∥ T x ∥ \geq c ∥ x ∥$ for all $x$ and some $c > 0$ ; $∥ A ∥ = \sup_{∥ x ∥ \leq 1} ∥ A x ∥$ is a bound for $A$ (A bounded operator that is bounded below, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

1.1F1F2algebra

Lower bound: for every $u \in H$ the coercivity of $a$ and the identity $a (u, u) = (A u, u)$ give $α ∥ u ∥^{2} \leq Re a (u, u) = Re (A u, u) \leq ∣ (A u, u) ∣ \leq ∥ A u ∥ ∥ u ∥ .$ If $u \neq 0$ divide by $∥ u ∥$ ; if $u = 0$ both sides vanish. Hence $α ∥ u ∥ \leq ∥ A u ∥$ for every $u \in H$ .

1.2F1F3

Upper bound: $∥ A u ∥ \leq ∥ A ∥ ∥ u ∥ \leq M ∥ u ∥$ for every $u$ , since $∥ A ∥ \leq M$ .

2.1F1F3step 1.1step 1.2∎

Consequences: by step 1.1, $A u = 0$ forces $α ∥ u ∥ \leq 0$ , hence $∥ u ∥ = 0$ and $u = 0$ , so $A$ is injective; together with step 1.1 this says exactly that $A$ is bounded below with constant $α$ , while step 1.2 supplies the upper bound, so $α ∥ u ∥ \leq ∥ A u ∥ \leq M ∥ u ∥$ for every $u \in H$ .

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

A bounded-below operator has closed range

Statement

Assume Countable Choice. Let $X$ be a Banach space, $Y$ a normed space over the same field, and let $T \in B (X, Y)$ satisfy $∥ T x ∥ \geq c ∥ x ∥$ for all $x \in X$ and some $c > 0$ (A bounded operator that is bounded below, A bounded linear operator between normed spaces). Then $T$ is injective and its range $ran T$ is a closed linear subspace of $Y$ ; the inverse $ran T \to X$ is bounded with norm at most $1 / c$ . The proof uses Countable Choice only to pass from sequential closedness to closedness; the Cauchy-sequence step uses completeness of $X$ .

Facts & Assumptions

Given: A Banach space $X$ , a normed space $Y$ over the same field, and a bounded linear operator $T : X \to Y$ with $∥ T x ∥ \geq c ∥ x ∥$ for all $x \in X$ , for a constant $c > 0$ ; write $Z : = ran T$ .

[F1]

Bounded below and bounded: $T$ is linear and bounded, and $∥ T x ∥ \geq c ∥ x ∥$ for every $x \in X$ ; also $T 0 = 0$ and $T (u + v) = T u + T v$ , $T (λ u) = λ T u$ (A bounded operator that is bounded below, A bounded linear operator between normed spaces).

[F2]

$X$ is complete: every Cauchy sequence in $X$ converges in $X$ (Banach space, Complete metric space: every Cauchy sequence converges in the space, Cauchy sequence in a metric space).

[F3]

In a metric space every sequentially closed set is closed, and this direction spends Countable Choice once, precisely by manufacturing a sequence from an adherence point (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

Limits in a metric space are unique (A sequence in a metric space has at most one limit).

[F5]

A subset $W$ of a vector space is a linear subspace exactly when $0 \in W$ , $u + v \in W$ and $λ u \in W$ for all $u, v \in W$ and all scalars $λ$ (Linear subspace of a vector space, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[F6]

Suppose $x_{k} \to x$ in $X$ and $C \geq 0$ is a bound for $T$ ; then $∥ T x_{k} - T x ∥ = ∥ T (x_{k} - x) ∥ \leq C ∥ x_{k} - x ∥ \to 0$ , so $T x_{k} \to T x$ : this is continuity of $T$ in the sequential and in the $ε$ - $δ$ forms (A bounded linear operator between normed spaces, Metric continuity characterisations, with countable choice for the sequential converse, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Vector addition and scalar multiplication are continuous in a normed space).

Proof

1.1F1

Injectivity: if $T x = 0$ then $0 = ∥ T x ∥ \geq c ∥ x ∥$ with $c > 0$ , so $∥ x ∥ = 0$ and $x = 0$ .

1.2F1F5

The range $Z$ is a linear subspace of $Y$ : $0 = T 0 \in Z$ ; if $z = T u$ and $w = T v$ then $z + w = T (u + v) \in Z$ ; and if $z = T u$ then $λ z = T (λ u) \in Z$ .

2.1F1F2F4F6step 1.1

Let $(z_{k}) \subseteq Z$ converge in $Y$ to some $y \in Y$ , say $z_{k} = T x_{k}$ with $x_{k}$ the unique preimage supplied by step 1.1. Then $∥ x_{k} - x_{m} ∥ \leq c^{- 1} ∥ T x_{k} - T x_{m} ∥ = c^{- 1} ∥ z_{k} - z_{m} ∥$ , so $(x_{k})$ is Cauchy in $X$ and hence converges to some $x \in X$ by completeness of $X$ . With any bound $C$ of $T$ , $∥ z_{k} - T x ∥ = ∥ T (x_{k} - x) ∥ \leq C ∥ x_{k} - x ∥ \to 0$ , so $z_{k} \to T x$ ; uniqueness of limits in $Y$ forces $y = T x \in Z$ . Thus $Z$ is sequentially closed in $Y$ .

2.2F1step 1.1algebra

Define $S : Z \to X$ by $S (y) : = x$ for the unique $x$ with $T x = y$ ; step 1.1 makes $S$ well defined with $T (S (y)) = y$ , and it is the inverse of $T$ viewed as a map onto $Z$ . For $y, z \in Z$ and scalars $a, b$ , applying $T$ to $a S (y) + b S (z)$ gives $a y + b z$ , so uniqueness gives $S (a y + b z) = a S (y) + b S (z)$ : the inverse is linear. For $y = T x \in Z$ we have $∥ S (y) ∥ = ∥ x ∥ \leq c^{- 1} ∥ T x ∥ = c^{- 1} ∥ y ∥$ , so $S$ is bounded with operator norm at most $1 / c$ .

3.1F3step 2.1

Since $Z$ is a sequentially closed subset of the metric space $Y$ , it is closed; this is the one step that uses Countable Choice, through the cited sequential-closure theorem.

4.1step 1.1step 1.2step 3.1step 2.2∎

Therefore $T$ is injective, $Z = ran T$ is a closed linear subspace of $Y$ , and the inverse map $S : Z \to X$ is bounded with norm at most $1 / c$ .

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

The adjoint of a coercive form is coercive with the same constants

Statement

Assume Countable Choice, used through the Riesz representation and Hilbert-adjoint suppliers. Let $a$ be a bounded sesquilinear form on a real or complex Hilbert space with bound $M$ and coercivity constant $α > 0$ (Bounded, coercive and symmetric sesquilinear forms), and let $a^{*} (u, v) : = \overline{a (v, u)}$ . Then $a^{*}$ is bounded with the same bound $M$ and coercive with the same constant $α$ ; its operator is the Hilbert adjoint $A^{*}$ of the operator $A$ of $a$ (A bounded form is represented by a unique bounded operator). In particular $\ker A^{*} = {0}$ and $(ran A)^{⊥} = \ker A^{*}$ , so the range of $A$ is dense in the classical route to surjectivity.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded sesquilinear form $a$ with bound $M \geq 0$ and coercivity constant $α > 0$ ; the adjoint form $a^{*} (u, v) = \overline{a (v, u)}$ ; and the operator $A$ of $a$ , $a (u, v) = (A u, v)$ .

[F1]

$a$ is linear in the first argument and conjugate-linear in the second, with $∣ a (u, v) ∣ \leq M ∥ u ∥ ∥ v ∥$ and $Re a (u, u) \geq α ∥ u ∥^{2}$ (Bounded, coercive and symmetric sesquilinear forms).

[F2]

The operator $A$ exists, is linear and bounded with $a (u, v) = (A u, v)$ , and $a^{*} (u, v) = (A^{*} u, v)$ where $A^{*}$ is the Hilbert adjoint of $A$ ; moreover the adjoint form $a^{*}$ is again a sesquilinear form (A bounded form is represented by a unique bounded operator, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F3]

A bounded coercive form's operator is bounded below with the coercivity constant: for the form $a^{*}$ with operator $A^{*}$ this gives $α ∥ u ∥ \leq ∥ A^{*} u ∥$ (A coercive form operator is bounded below).

[F4]

Kernel--range orthogonality: $(ran A)^{⊥} = \ker A^{*}$ and $\overline{ran A} = (\ker A^{*})^{⊥}$ for the Hilbert adjoint (Kernel–range orthogonality for Hilbert adjoints, The Hilbert-space adjoint of a bounded operator).

[F5]

Conjugation is an involution with $Re \overline{z} = Re z$ and $∣ z ∣ = ∣ \overline{z} ∣$ (Real and imaginary parts, complex conjugation, and modulus, Real and complex inner-product spaces and their induced length).

Proof

1.1F1F2F5

Boundedness of $a^{*}$ : for all $u, v \in H$ , $∣ a^{*} (u, v) ∣ = ∣ \overline{a (v, u)} ∣ = ∣ a (v, u) ∣ \leq M ∥ v ∥ ∥ u ∥$ , so $a^{*}$ is bounded with the same bound $M$ ; it is sesquilinear of the same type, being conjugate-linear in $v$ and linear in $u$ .

1.2F1F5algebra

Coercivity of $a^{*}$ : $a^{*} (u, u) = \overline{a (u, u)}$ has the same real part as $a (u, u)$ , hence $Re a^{*} (u, u) = Re a (u, u) \geq α ∥ u ∥^{2}$ for every $u$ .

2.1F2F3step 1.2

Operator and kernel: [F2] identifies the operator of $a^{*}$ as the Hilbert adjoint $A^{*}$ ; since $a^{*}$ is bounded and coercive with constant $α$ , [F3] gives $α ∥ u ∥ \leq ∥ A^{*} u ∥$ , so $A^{*} u = 0$ forces $u = 0$ , that is $\ker A^{*} = {0}$ .

3.1F4step 2.1∎

Orthogonality: by [F4], $(ran A)^{⊥} = \ker A^{*} = {0}$ , so the orthogonal complement of the range of $A$ is trivial and $\overline{ran A} = (\ker A^{*})^{⊥} = H$ ; the range of $A$ is dense.

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

Coercivity of the adjoint makes the form-operator range dense

Statement

Assume Countable Choice. Let $a$ be a bounded coercive sesquilinear form on a real or complex Hilbert space $H$ with constants $M, α$ , and let $A$ be the operator with $a (u, v) = (A u, v)$ (A bounded form is represented by a unique bounded operator). Then $(ran A)^{⊥} = \ker A^{*} = {0}, \overline{ran A} = H .$ Combined with the closedness from A bounded-below operator has closed range this gives $ran A = H$ ; this is the classical closed-range/density route to Lax--Milgram, recorded here as the pla's operator-level density step.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ with constants $M, α$ ; its operator $A \in B (H)$ with $a (u, v) = (A u, v)$ ; and the adjoint form $a^{*}$ with operator $A^{*}$ .

[F1]

The adjoint form $a^{*} (u, v) = \overline{a (v, u)}$ is bounded with bound $M$ and coercive with the same constant $α$ , and its operator is the Hilbert adjoint $A^{*}$ ; also $a$ coercive with constant $α$ makes $A$ bounded below with constant $α$ (The adjoint of a coercive form is coercive with the same constants, A bounded form is represented by a unique bounded operator, A coercive form operator is bounded below, The Hilbert-space adjoint of a bounded operator).

[F2]

Orthogonal complements: $(ran A)^{⊥} = \ker A^{*}$ , and for every linear subspace $M$ of $H$ one has $M^{⊥ ⊥} = \overline{M}$ , with ${0}^{⊥} = H$ (Kernel–range orthogonality for Hilbert adjoints, The double orthogonal complement of a subspace is its closure, Orthogonality and the orthogonal complement).

[F3]

A bounded-below operator on a Banach space has closed range: applied to $A : H \to H$ , whose domain $H$ is complete, this gives that $ran A$ is closed (A bounded-below operator has closed range).

[F4]

Coercivity of $a$ with constant $α$ means $Re a (u, u) \geq α ∥ u ∥^{2}$ for all $u$ (Bounded, coercive and symmetric sesquilinear forms).

Proof

1.1F1

$\ker A^{*} = {0}$ : by [F1] the form $a^{*}$ is bounded and coercive with constant $α$ and has operator $A^{*}$ , so $A^{*}$ is bounded below with constant $α$ ; hence $A^{*} u = 0$ forces $α ∥ u ∥ \leq 0$ and $u = 0$ .

2.1F2step 1.1

Density: by [F2], $(ran A)^{⊥} = \ker A^{*} = {0}$ , and the double orthogonal complement theorem applied to the linear subspace $ran A$ gives $\overline{ran A} = (ran A)^{⊥ ⊥} = {0}^{⊥} = H$ . So the range of $A$ is dense in $H$ .

3.1F3F4step 2.1∎

Closedness and surjectivity: by [F1] and [F4], $A$ is bounded below with constant $α$ ; since $H$ is complete, [F3] makes $ran A$ closed. A dense closed subset of a metric space is the whole space, so $ran A = H$ ; combined with step 2.1 this is the classical closed-range/density route to surjectivity of $A$ .

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

Coercivity makes a small form step a strict contraction

Statement

Assume Countable Choice, used through A bounded form is represented by a unique bounded operator. Let $H \neq {0}$ be a real or complex Hilbert space and let $a$ be a bounded coercive sesquilinear form on $H$ with constants $M > 0, α > 0$ (so necessarily $α \leq M$ ); let $A$ be its operator. Put $q_{ρ} : = \sqrt{1 - 2 ρ α + ρ^{2} M^{2}}$ for $0 < ρ < 2 α / M^{2}$ . Then $0 \leq q_{ρ} < 1$ , and for every $w \in H$ the map $T_{ρ, w} (u) : = u - ρ (A u - w)$ is a strict contraction of $H$ with constant $q_{ρ}$ : $∥ T_{ρ, w} (u) - T_{ρ, w} (v) ∥ \leq q_{ρ} ∥ u - v ∥$ for all $u, v$ . In particular $I - ρ A$ is a strict contraction with the same constant. The estimate is the only place where the coercivity constant and the bound enter the contraction argument.

Facts & Assumptions

Given: Countable Choice; a Hilbert space $H \neq {0}$ with inner product linear in the first argument; a bounded coercive sesquilinear form $a$ with constants $M > 0, α > 0$ ; its operator $A$ , $a (u, v) = (A u, v)$ ; and a real $ρ$ with $0 < ρ < 2 α / M^{2}$ .

[F1]

$A$ is linear with $a (u, v) = (A u, v)$ for all $u, v$ ; $Re a (u, u) = Re (A u, u) \geq α ∥ u ∥^{2}$ and $∥ A u ∥ \leq M ∥ u ∥$ (Bounded, coercive and symmetric sesquilinear forms, A bounded form is represented by a unique bounded operator, A coercive form operator is bounded below).

[F2]

Inner-product norm expansion: $∥ z - ρ A z ∥^{2} = ∥ z ∥^{2} - 2 ρ Re (A z, z) + ρ^{2} ∥ A z ∥^{2}$ , and $Re (A z, z) \leq ∣ (A z, z) ∣ \leq ∥ A z ∥ ∥ z ∥$ by Cauchy--Schwarz; the induced length is a norm with the triangle inequality (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, The induced length is a norm, Real and imaginary parts, complex conjugation, and modulus).

[F3]

A map $S : H \to H$ is a strict contraction with constant $q$ when $∥ S (u) - S (v) ∥ \leq q ∥ u - v ∥$ for all $u, v$ and $0 \leq q < 1$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[F4]

Nonnegative square roots: for $t \geq 0$ there is a unique $s \geq 0$ with $s^{2} = t$ , denoted $\sqrt{t}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). If $0 \leq x \leq y$ , then $\sqrt{x} > \sqrt{y}$ would imply $x - y = (\sqrt{x} - \sqrt{y}) (\sqrt{x} + \sqrt{y}) > 0$ , a contradiction; hence $\sqrt{x} \leq \sqrt{y}$ .

Proof

1.1F1algebra

On a nonzero Hilbert space the named constants satisfy $α \leq M$ : choosing $u \neq 0$ and dividing $u$ by its norm, $α \leq Re a (u, u) \leq ∣ a (u, u) ∣ \leq M ∥ u ∥^{2} = M$ , The identity $1 - 2 ρ α + ρ^{2} M^{2} = M^{2} (ρ - α / M^{2})^{2} + 1 - α^{2} / M^{2}$ makes the radicand nonnegative; it may vanish when $α = M$ and $ρ = α / M^{2}$ .

1.2F1F2algebra

Contraction estimate: for $z \in H$ and $0 < ρ < 2 α / M^{2}$ the expansion of [F2] together with [F1] gives $∥ z - ρ A z ∥^{2} = ∥ z ∥^{2} - 2 ρ Re (A z, z) + ρ^{2} ∥ A z ∥^{2} \leq (1 - 2 ρ α + ρ^{2} M^{2}) ∥ z ∥^{2} = q_{ρ}^{2} ∥ z ∥^{2},$ so $∥ (I - ρ A) z ∥ \leq q_{ρ} ∥ z ∥$ ; taking $z = u - v$ and using $T_{ρ, w} (u) - T_{ρ, w} (v) = (I - ρ A) (u - v)$ gives the contraction estimate for every $w$ .

2.1F3F4step 1.1step 1.2algebra∎

The constant lies in $[0, 1)$ : the radicand is a quadratic in $ρ$ with minimum $1 - α^{2} / M^{2}$ at $ρ = α / M^{2}$ , which is nonnegative because $α \leq M$ by step 1.1; at the endpoints $ρ = 0$ and $ρ = 2 α / M^{2}$ it equals $1$ , and for $0 < ρ < 2 α / M^{2}$ either directly $1 - 2 ρ α + ρ^{2} M^{2} < 1$ or by the strict minimum unless $ρ = α / M^{2}$ and $α = M$ , in which case the radicand vanishes and $q_{ρ} = 0$ ; in every case $0 \leq q_{ρ} < 1$ . Hence $I - ρ A$ , and with it $T_{ρ, w}$ , is a strict contraction with constant $q_{ρ}$ .

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The Lax--Milgram theorem

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real or complex Hilbert space, let $a$ be a bounded coercive sesquilinear form on $H$ with constants $M, α$ (Bounded, coercive and symmetric sesquilinear forms), and let $F : H \to K$ be a bounded conjugate-linear functional with norm $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ . Then there is a unique $u \in H$ with $a (u, v) = F (v) for every v \in H,$ and it satisfies $α ∥ u ∥ \leq ∥ F ∥$ , that is $∥ u ∥ \leq ∥ F ∥ / α$ . The real bilinear case is the same statement with $a$ symmetric or not, $F$ a bounded linear functional, and the conjugation read as the identity.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ on $H$ with constants $M \geq 0$ and $α > 0$ ; a bounded conjugate-linear functional $F$ on $H$ with $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ ; and the operator $A$ of $a$ , $a (u, v) = (A u, v)$ .

[F1]

$A$ is linear and bounded with $a (u, v) = (A u, v)$ , $∥ A ∥ \leq M$ , and $Re a (u, u) \geq α ∥ u ∥^{2}$ (Bounded, coercive and symmetric sesquilinear forms, A bounded form is represented by a unique bounded operator, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F2]

Riesz representation under Countable Choice: every bounded linear functional $G$ has a unique $w$ with $G (v) = (v, w)$ for all $v$ , and $∥ G ∥ = ∥ w ∥$ (Riesz representation for Hilbert spaces, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

Contractions on complete spaces: a map $T$ of a nonempty complete metric space with $∥ T (u) - T (v) ∥ \leq q ∥ u - v ∥$ , $0 \leq q < 1$ , has exactly one fixed point; and for $H \neq {0}$ and $0 < ρ < 2 α / M^{2}$ the map $u \mapsto u - ρ (A u - w)$ is a strict contraction with constant $q_{ρ} = (1 - 2 ρ α + ρ^{2} M^{2})^{1 / 2} < 1$ (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Coercivity makes a small form step a strict contraction, Hilbert space).

[F4]

Conjugation: $G (v) : = \overline{F (v)}$ is linear when $F$ is conjugate-linear, $∣ G (v) ∣ = ∣ F (v) ∣$ , and $G$ bounded with the same norm; $Re z \leq ∣ z ∣$ and $∣ z ∣ = ∣ \overline{z} ∣$ (Real and imaginary parts, complex conjugation, and modulus).

[F5]

Linearity of $a$ in the first argument and the estimate for the unique solution follow from $a (u, v) = (A u, v)$ ; the degenerate space $H = {0}$ has the unique solution $u = 0$ and $∥ F ∥ = 0$ (Bounded, coercive and symmetric sesquilinear forms, Hilbert space).

Proof

1.1F1algebra

Assume first $H \neq {0}$ . Then the given bound and coercivity constant satisfy $α \leq M$ and $M > 0$ : choosing $u \neq 0$ and normalising, $α \leq Re a (u, u) \leq ∣ a (u, u) ∣ \leq M ∥ u ∥^{2}$ , so $M \geq α > 0$ ; hence $ρ : = α / M^{2}$ satisfies $0 < ρ < 2 α / M^{2}$ .

1.2F1algebra

Uniqueness: if $u$ satisfies $a (u, v) = 0$ for every $v$ , then testing $v = u$ gives $α ∥ u ∥^{2} \leq Re a (u, u) = 0$ , so $u = 0$ . If $u_{1}, u_{2}$ are two solutions of $a (u, \cdot) = F$ , then first-slot linearity gives $a (u_{1} - u_{2}, v) = 0$ for all $v$ , so $u_{1} = u_{2}$ .

2.1F2F4step 1.1

The conjugate functional: $G (v) : = \overline{F (v)}$ is a bounded linear functional with $∥ G ∥ = ∥ F ∥$ , so by Riesz representation there is a unique $w \in H$ with $G (v) = (v, w)$ for all $v$ , that is $F (v) = (w, v)$ for all $v$ , and $∥ w ∥ = ∥ F ∥$ .

3.1F1F3step 2.1

Existence: fix $ρ$ as in step 1.1 and define $T (u) : = u - ρ (A u - w)$ . By [F3] the map $T$ is a strict contraction of the complete space $H$ with constant $q_{ρ} < 1$ , so by the Banach fixed-point theorem it has a fixed point $u \in H$ ; then $u = u - ρ (A u - w)$ gives $A u = w$ , and hence $a (u, v) = (A u, v) = (w, v) = F (v)$ for every $v \in H$ .

4.1F1F4step 3.1algebra

Estimate: for the solution $u$ of step 3.1, $α ∥ u ∥^{2} \leq Re a (u, u) = Re F (u) \leq ∣ F (u) ∣ \leq ∥ F ∥ ∥ u ∥$ ; if $u \neq 0$ divide by $∥ u ∥$ to get $α ∥ u ∥ \leq ∥ F ∥$ , and if $u = 0$ the inequality holds trivially.

5.1F5given∎

Degenerate space: if $H = {0}$ , then the only element is $0$ , the only functional is $0$ and it is the value at the unique solution $u = 0$ with $α ∥ u ∥ = 0 = ∥ F ∥$ ; uniqueness is immediate. The real bilinear case is the same argument with conjugation read as the identity, and $a$ need not be symmetric.

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The Lax--Milgram solution operator has norm at most $1 / α$

Statement

Under the hypotheses of The Lax--Milgram theorem, let $F$ be the space of bounded conjugate-linear functionals on $H$ , normed by $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ , and let $S : F \to H$ assign to $F$ the unique solution $u$ of $a (u, v) = F (v)$ for all $v$ . Then $S$ is well defined and linear, and $∥ S ∥ \leq \frac{1}{α} .$ In particular $∥ u ∥ \leq ∥ F ∥ / α$ for every $F$ , and the estimate is uniform over all data.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ with constants $M, α$ ; the normed space $F$ of bounded conjugate-linear functionals on $H$ with $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ ; and the solution map $S : F \to H$ sending $F$ to the unique $u$ with $a (u, v) = F (v)$ for all $v$ .

[F1]

Lax--Milgram: for every $F \in F$ there is exactly one $u \in H$ with $a (u, v) = F (v)$ for all $v$ , and $α ∥ u ∥ \leq ∥ F ∥$ ; the form is linear in the first argument (The Lax--Milgram theorem, Bounded, coercive and symmetric sesquilinear forms, Hilbert space).

[F2]

Bounded operators and operator norm: $∥ S ∥ = \sup {∥ S F ∥ : ∥ F ∥ \leq 1}$ , and $S$ is bounded with $∥ S ∥ \leq 1 / α$ once $∥ S F ∥ \leq ∥ F ∥ / α$ for every $F$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

1.1F1

$S$ is well defined by [F1]: each $F$ has exactly one solution, so $S$ is a function $F \to H$ .

2.1F1step 1.1algebra

$S$ is linear: if $u_{i} = S (F_{i})$ and $λ$ is a scalar, then for every $v$ , first-slot linearity of $a$ gives $a (u_{1} + u_{2}, v) = F_{1} (v) + F_{2} (v)$ and $a (λ u_{1}, v) = λ F_{1} (v)$ ; by the uniqueness part of [F1], $S (F_{1} + F_{2}) = S (F_{1}) + S (F_{2})$ and $S (λ F_{1}) = λ S (F_{1})$ .

3.1F1F2step 2.1∎

Norm bound: the estimate of [F1] reads $∥ S (F) ∥ \leq ∥ F ∥ / α$ for every $F \in F$ ; hence $S$ is a bounded linear operator with $∥ S ∥ \leq 1 / α$ by [F2]. The same inequality gives $∥ u ∥ \leq ∥ F ∥ / α$ for each datum, uniformly.

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Symmetric Lax--Milgram is energy minimisation

Statement

Assume Countable Choice. Let $a$ be a bounded coercive symmetric sesquilinear form on a real or complex Hilbert space $H$ with coercivity constant $α > 0$ and let $F$ be a bounded conjugate-linear functional, with $u$ the Lax--Milgram solution of The Lax--Milgram theorem. Then the functional $J (v) : = \frac{1}{2} Re a (v, v) - Re F (v)$ is real valued and attains its strict minimum on $H$ at $u$ : $J (v) > J (u)$ for every $v \neq u$ . In the real case $J (v) = \frac{1}{2} a (v, v) - F (v)$ , and in the complex case $a (v, v)$ is already real by symmetry, so $J (v) = \frac{1}{2} a (v, v) - Re F (v)$ ; no claim is made that a nonsymmetric form has such a minimisation. Consequently the solution is characterised by the minimisation problem independently of uniqueness in The Lax--Milgram theorem.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive symmetric sesquilinear form $a$ with coercivity constant $α > 0$ ; a bounded conjugate-linear functional $F$ ; the Lax--Milgram solution $u$ with $a (u, v) = F (v)$ for all $v$ ; and $J (v) = \frac{1}{2} Re a (v, v) - Re F (v)$ .

[F1]

Symmetry makes $a (v, v)$ real: $a (v, v) = \overline{a (v, v)}$ , so $Re a (v, v) = a (v, v)$ ; coercivity gives $Re a (w, w) \geq α ∥ w ∥^{2}$ ; and $a$ is linear in the first argument and conjugate-linear in the second, so $a$ is additive in each slot and $a (w, u) = \overline{a (u, w)}$ by symmetry (Bounded, coercive and symmetric sesquilinear forms, The induced length is a norm).

[F2]

The solution satisfies $a (u, w) = F (w)$ for every $w \in H$ ; existence and uniqueness are those of Lax--Milgram (The Lax--Milgram theorem).

[F3]

Real parts: $Re z + Re \overline{z} = 2 Re z$ and $Re (z_{1} + z_{2}) = Re z_{1} + Re z_{2}$ (Real and imaginary parts, complex conjugation, and modulus, Hilbert space).

Proof

1.1F1F3algebra

Expansion: write $v = u + w$ with $w : = v - u$ . Using additivity in both slots, symmetry and [F3], $Re a (v, v) = Re a (u, u) + Re (a (u, w) + \overline{a (u, w)}) + Re a (w, w) = Re a (u, u) + 2 Re a (u, w) + Re a (w, w),$ while $Re F (v) = Re F (u) + Re F (w)$ , so $J (v) - J (u) = \frac{1}{2} Re a (w, w) + Re a (u, w) - Re F (w) .$

2.1F1F2step 1.1

The linear term vanishes: by [F2], $a (u, w) = F (w)$ , hence $Re a (u, w) - Re F (w) = 0$ , and $J (v) - J (u) = \frac{1}{2} Re a (w, w) \geq \frac{α}{2} ∥ w ∥^{2}$ by coercivity.

3.1F1F3step 2.1∎

Strict minimum: the right-hand side is positive whenever $w \neq 0$ ; hence $J (v) > J (u)$ for every $v \neq u$ , so $J$ is real valued and attains its strict minimum at the Lax--Milgram solution $u$ . In the real case $Re$ is the identity and $J (v) = \frac{1}{2} a (v, v) - F (v)$ ; in the complex case symmetry makes $a (v, v)$ real, so taking its real part is redundant, while $Re F (v)$ ensures a real-valued functional. No minimisation claim is made for nonsymmetric forms.

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Nonsymmetric Lax--Milgram is not a scalar minimisation principle

Remark

Existence and uniqueness in The Lax--Milgram theorem do not require symmetry: only boundedness and coercivity are used, and the contraction proof never symmetrises the form. Symmetry is, however, exactly what the energy characterisation of Symmetric Lax--Milgram is energy minimisation consumes. If $a$ is bounded and coercive but not symmetric, and $F$ is bounded and conjugate-linear, then the solution of $a (u, v) = F (v)$ is in general not a critical point, and not a minimiser, of $v \mapsto \frac{1}{2} Re a (v, v) - Re F (v)$ : the Euler--Lagrange equation of that functional involves the symmetrised form $a_{s} = \frac{1}{2} (a + a^{*})$ , whose operator is $\frac{1}{2} (A + A^{*})$ , whereas the weak equation involves $A$ . The two coincide exactly when $a$ is symmetric. The companion page's positive-definite-plus-skew counterexample and drift example exhibit the failure concretely. This is a remark: it records the boundary of the variational statement and is not used as a proof step.

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The negative Sobolev space $H^{- 1} (Ω)$

Definition

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $Ω \subseteq R^{n}$ be open, $n \geq 1$ , with $H_{0}^{1} (Ω; K)$ the zero-boundary Sobolev space $W_{0}^{1, 2}$ over $K \in {R, C}$ (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol). Define $H^{- 1} (Ω) : = {F : H_{0}^{1} (Ω) \to K : F is bounded and conjugate-linear},$ with $∥ F ∥_{H^{- 1}} : = \sup_{∥ v ∥_{H_{0}^{1}} \leq 1} ∣ F (v) ∣ .$ Pairing convention. The pairing $⟨ F, v ⟩ : = F (v)$ is linear in $F$ and conjugate-linear in $v$ ; it is the dual pairing of $H^{- 1} (Ω)$ with $H_{0}^{1} (Ω)$ , not the $L^{2}$ inner product. The map $F \mapsto \overline{F (\cdot)}$ is an isometric conjugate-linear bijection of $H^{- 1} (Ω)$ onto the Banach dual $(H_{0}^{1} (Ω))^{*}$ of The dual space X^* of a normed space and its dual norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators); the design's notation $(H_{0}^{1})^{*}$ is read through this identification, which is the one compatible with the page's sesquilinear convention (linear in the first argument, conjugate-linear in the second) fixed in Bounded, coercive and symmetric sesquilinear forms. The space $H^{- 1} (Ω)$ is a normed space, complete because $(H_{0}^{1} (Ω))^{*}$ is complete and $\overline{F (\cdot)}$ is an isometry in both directions (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, Real and imaginary parts, complex conjugation, and modulus). For $f \in L^{2} (Ω)$ the map $v \mapsto (f, v)_{L^{2}}$ is the corresponding element of $H^{- 1} (Ω)$ under the conventions of The space $L^{p} (μ)$ as the quotient by null functions and Complex Lp classes and Euclidean test-function conventions, where complex integrability is in the sense of the latter; identifying a general element of $H^{- 1}$ with an $L^{2}$ function is an embedding statement, never a definition. Elements of $H^{- 1}$ are defined here by their action on Sobolev classes; this does not exclude their identification with distributions through smooth test functions.

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$L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound

Statement

Assume the Axiom of Choice, inherited through the Poincar'e supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, nonempty, bounded in one direction (so the Poincar'e inequality of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction holds; every bounded open set qualifies), and let $f_{0}, f_{1}, \dots, f_{n} \in L^{2} (Ω; K)$ . Define $F (v) : = (f_{0}, v)_{L^{2}} + \sum_{i = 1}^{n} (f_{i}, D_{i} v)_{L^{2}}, v \in H_{0}^{1} (Ω),$ with the $L^{2}$ inner product of $L^{2}$ with the integral pairing is a Hilbert space. Then $F$ is a well-defined conjugate-linear functional on $H_{0}^{1} (Ω)$ , independent of the $L^{2}$ classes chosen only through those classes, and bounded: with the Poincar'e constant $C_{P}$ of $Ω$ for $W_{0}^{1, 2}$ , $∣ F (v) ∣ \leq (C_{P} ∥ f_{0} ∥_{L^{2}} + \sum_{i = 1}^{n} ∥ f_{i} ∥_{L^{2}}) ∥ v ∥_{H_{0}^{1}}, ∥ F ∥_{H^{- 1}} \leq C_{P} ∥ f_{0} ∥_{L^{2}} + \sum_{i = 1}^{n} ∥ f_{i} ∥_{L^{2}} .$ In particular $F \in H^{- 1} (Ω)$ in the sense of The negative Sobolev space $H^{- 1} (Ω)$ , and if $Ω$ is bounded then every $f \in L^{2} (Ω)$ defines an $H^{- 1}$ element by $v \mapsto (f, v)_{L^{2}}$ . The functional is the weak form of $f_{0} - \sum_{i} D_{i} f_{i}$ ; no claim that every $H^{- 1}$ element arises this way is made here.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open, nonempty $Ω \subseteq R^{n}$ bounded in one direction, with a unit vector $e$ and reals $a < b$ such that $a < x \cdot e < b$ for all $x \in Ω$ ; classes $f_{0}, f_{1}, \dots, f_{n} \in L^{2} (Ω; K)$ ; and the functional $F (v) = (f_{0}, v)_{L^{2}} + \sum_{i = 1}^{n} (f_{i}, D_{i} v)_{L^{2}}$ on $H_{0}^{1} (Ω)$ .

[F1]

$H^{- 1} (Ω)$ is the space of bounded conjugate-linear functionals on $H_{0}^{1} (Ω)$ with $∥ F ∥_{H^{- 1}} = \sup_{∥ v ∥_{H_{0}^{1}} \leq 1} ∣ F (v) ∣$ ; the pairing $⟨ F, v ⟩ = F (v)$ is linear in $F$ and conjugate-linear in $v$ (The negative Sobolev space $H^{- 1} (Ω)$ , The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F2]

$H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω; K)$ carries the $W^{1, 2}$ norm, for which $∥ v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ , $∥ D_{i} v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ , and $∥ D v ∥_{L^{2}} = (\sum_{i = 1}^{n} ∥ D_{i} v ∥_{L^{2}}^{2})^{1 / 2}$ satisfies $∥ D v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ ; the weak derivatives are class operators in the sense of Weak derivative of a locally integrable function (Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

Poincar'e inequality: with $C_{P} : = C (2) (b - a)$ for the constant of the cited theorem, $∥ v ∥_{L^{2} (Ω)} \leq C_{P} ∥ D v ∥_{L^{2} (Ω)}$ for every $v \in H_{0}^{1} (Ω)$ . This is the claim of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction at $p = 2$ , stated there for $W_{0}^{1, p}$ classes.

[F4]

H"older and the $L^{2}$ pairing: $∣ (f, v)_{L^{2}} ∣ \leq ∥ f ∥_{L^{2}} ∥ v ∥_{L^{2}}$ for classes, the pairing is conjugate-linear in its second argument, and it depends only on the two classes (Holder's inequality for integrals, including the endpoint cases, The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions).

[F5]

The Axiom of Choice supplies Countable Choice for the Sobolev and $L^{2}$ interfaces (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F6]

$L^{2}$ functions are locally integrable by H"older on compact sets; they define regular distributions, whose coordinate derivatives satisfy $⟨ \partial_{i} u_{f}, φ ⟩ = - \int f D_{i} φ$ (Locally integrable functions embed in distributions, Regular distribution from a locally integrable function, Distributional derivative).

Proof

1.1F2F4

$F$ is well defined and conjugate-linear. Each summand $v \mapsto (f_{i}, D_{i} v)_{L^{2}}$ is a composition of the class map $v \mapsto D_{i} v$ , which is linear on Sobolev classes, with the $L^{2}$ pairing, which is conjugate-linear in its second argument; hence each summand is conjugate-linear and depends only on the class of $v$ and the class of $f_{i}$ . A finite sum of conjugate-linear functionals is conjugate-linear, so $F$ is a well-defined conjugate-linear functional on $H_{0}^{1} (Ω)$ .

2.1F1F2F3F4F5step 1.1algebra

Bound. For every $v \in H_{0}^{1} (Ω)$ , H"older gives $∣ (f_{0}, v)_{L^{2}} ∣ \leq ∥ f_{0} ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq C_{P} ∥ f_{0} ∥_{L^{2}} ∥ D v ∥_{L^{2}}$ and $∣ (f_{i}, D_{i} v)_{L^{2}} ∣ \leq ∥ f_{i} ∥_{L^{2}} ∥ D_{i} v ∥_{L^{2}}$ for each $i$ ; summing and using $∥ D v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ and $∥ D_{i} v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ gives $∣ F (v) ∣ \leq (C_{P} ∥ f_{0} ∥_{L^{2}} + \sum_{i = 1}^{n} ∥ f_{i} ∥_{L^{2}}) ∥ v ∥_{H_{0}^{1}} .$ Consequently $F$ is bounded with $∥ F ∥_{H^{- 1}} \leq C_{P} ∥ f_{0} ∥_{L^{2}} + \sum_{i} ∥ f_{i} ∥_{L^{2}}$ , so $F \in H^{- 1} (Ω)$ .

3.1F3step 2.1

The pure $L^{2}$ case: if $Ω$ is bounded, then it is bounded in one direction --- for any unit vector $e$ and any $r$ with $Ω \subseteq B (0, r)$ one has $- r < x \cdot e < r$ --- so the hypothesis holds and $F (v) : = (f_{0}, v)_{L^{2}}$ with $f_{1} = \dots = f_{n} = 0$ is an element of $H^{- 1} (Ω)$ with $∥ F ∥_{H^{- 1}} \leq C_{P} ∥ f_{0} ∥_{L^{2}}$ .

4.1F2F4F6step 2.1∎

Identification with the divergence-form datum: let $u_{f_{i}}$ be the regular distribution associated to $f_{i} \in L^{2} \subset L_{l o c}^{1}$ and put $T : = u_{f_{0}} - \sum_{i} \partial_{i} u_{f_{i}}$ . For $v \in C_{c}^{\infty} (Ω)$ , the definition of the distributional derivative gives $⟨ T, \overline{v} ⟩ = \int f_{0} \overline{v} + \sum_{i} \int f_{i} \overline{D_{i} v} = F (v)$ . Thus $F$ extends this conjugated test pairing boundedly to $H_{0}^{1} (Ω)$ ; no function-valued derivative of any $f_{i}$ is assumed. The representation by data $(f_{0}, \dots, f_{n})$ is not asserted to be unique and no surjectivity onto $H^{- 1} (Ω)$ is claimed.

Remarks

The converse representation is proved in Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence.

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

Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, $n \geq 1$ . For every $F \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ) there are $f_{0}, f_{1}, \dots, f_{n} \in L^{2} (Ω)$ such that $F (v) = (f_{0}, v)_{L^{2}} + \sum_{i = 1}^{n} (f_{i}, D_{i} v)_{L^{2}} for every v \in H_{0}^{1} (Ω),$ the norm is exactly the infimum over all such representations, $∥ F ∥_{H^{- 1}} = \inf {(\sum_{i = 0}^{n} ∥ f_{i} ∥_{L^{2}}^{2})^{1 / 2} : F (v) = (f_{0}, v)_{L^{2}} + \sum_{i = 1}^{n} (f_{i}, D_{i} v)_{L^{2}} \forall v},$ and the infimum is attained by the canonical choice $f_{0} = g$ , $f_{i} = D_{i} g$ given by the Riesz vector of $\overline{F (\cdot)}$ ; in particular the data are controlled by the norm and conversely. The representation is the converse of $L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound and needs no Hahn--Banach extension theorem: Riesz representation in $H_{0}^{1}$ already produces it.

Facts & Assumptions

Given: Countable Choice; an open $Ω \subseteq R^{n}$ , $n \geq 1$ ; and a functional $F \in H^{- 1} (Ω)$ , that is, a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ with $∥ F ∥_{H^{- 1}} = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ .

[F1]

$H_{0}^{1} (Ω)$ is a Hilbert space under $(u, v)_{H^{1}} : = (u, v)_{L^{2}} + \sum_{i = 1}^{n} (D_{i} u, D_{i} v)_{L^{2}}$ , whose induced norm is the $W^{1, 2}$ norm; the pairing is linear in the first variable and conjugate-linear in the second, and conjugate-symmetric (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure, The notation $H^{k}$ and the reserved zero-boundary symbol, Hilbert space).

[F2]

$H^{- 1} (Ω)$ consists of the bounded conjugate-linear functionals, with $∥ F ∥_{H^{- 1}} = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ ; the $L^{2}$ pairings are conjugate-symmetric and depend only on classes (The negative Sobolev space $H^{- 1} (Ω)$ , The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

Riesz representation under Countable Choice: for a bounded linear functional $G$ on $H_{0}^{1} (Ω)$ there is a unique $g$ with $G (v) = (v, g)_{H^{1}}$ for all $v$ , and $∥ G ∥ = ∥ g ∥_{H^{1}}$ , the operator norm being the dual norm (Riesz representation for Hilbert spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

Every $L^{2}$ pairing satisfies $∣ (f, v)_{L^{2}} ∣ \leq ∥ f ∥_{2} ∥ v ∥_{2}$ by $L^{2}$ with the integral pairing is a Hilbert space and Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs. Conjugation and finite Cauchy--Schwarz: $\overline{z}$ has $∣ \overline{z} ∣ = ∣ z ∣$ and $\overline{F (v)}$ depends linearly on $v$ when $F$ is conjugate-linear; for complex numbers $z_{0}, \dots, z_{n}, w_{0}, \dots, w_{n}$ , $∣ \sum_{i = 0}^{n} z_{i} \overline{w_{i}} ∣ \leq (\sum_{i = 0}^{n} ∣ z_{i} ∣^{2})^{1 / 2} (\sum_{i = 0}^{n} ∣ w_{i} ∣^{2})^{1 / 2}$ (Real and imaginary parts, complex conjugation, and modulus, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

1.1F1F4

The functional $G (v) : = \overline{F (v)}$ is linear in $v$ (conjugating a conjugate-linear map gives a linear one), and $∣ G (v) ∣ = ∣ F (v) ∣$ for every $v$ , so $G$ is bounded with the same dual norm as $F$ .

1.2F1F2F4

Every representation bounds the norm: if $F (v) = (f_{0}, v)_{L^{2}} + \sum_{i} (f_{i}, D_{i} v)_{L^{2}}$ for all $v$ , then the $L^{2}$ pairing bound followed by finite Cauchy--Schwarz on the real vectors of component norms, and $∥ v ∥_{H^{1}}^{2} = ∥ v ∥_{L^{2}}^{2} + \sum_{i} ∥ D_{i} v ∥_{L^{2}}^{2}$ give $∣ F (v) ∣ \leq (\sum_{i = 0}^{n} ∥ f_{i} ∥_{L^{2}}^{2})^{1 / 2} ∥ v ∥_{H^{1}},$ so $∥ F ∥_{H^{- 1}} \leq (\sum_{i = 0}^{n} ∥ f_{i} ∥_{L^{2}}^{2})^{1 / 2}$ and, taking the infimum over all representations, $∥ F ∥_{H^{- 1}} \leq \inf {\dots}$ .

2.1F1F2F3F4step 1.1

Riesz representation: by [F3] applied to the Hilbert space $H_{0}^{1} (Ω)$ there is a unique $g \in H_{0}^{1} (Ω)$ with $G (v) = (v, g)_{H^{1}}$ for every $v \in H_{0}^{1} (Ω)$ , and $∥ g ∥_{H^{1}} = ∥ G ∥ = ∥ F ∥_{H^{- 1}}$ . Conjugating and expanding the $H^{1}$ inner product gives, for every $v$ , $F (v) = \overline{(v, g)_{H^{1}}} = (g, v)_{L^{2}} + \sum_{i = 1}^{n} (D_{i} g, D_{i} v)_{L^{2}},$ so with $f_{0} : = g$ and $f_{i} : = D_{i} g \in L^{2} (Ω)$ this is a representation of the required form.

3.1F1F3step 2.1step 1.2algebra∎

The canonical representation attains the infimum: for $f_{0} = g$ , $f_{i} = D_{i} g$ one has $\sum_{i = 0}^{n} ∥ f_{i} ∥_{L^{2}}^{2} = ∥ g ∥_{L^{2}}^{2} + \sum_{i} ∥ D_{i} g ∥_{L^{2}}^{2} = ∥ g ∥_{H^{1}}^{2} = ∥ F ∥_{H^{- 1}}^{2}$ by step 2.1, so the infimum is at most $∥ F ∥_{H^{- 1}}$ and, with step 1.2, equals it. The data of the canonical representation are controlled by the norm through $∥ g ∥_{H^{1}} = ∥ F ∥_{H^{- 1}}$ and conversely by the estimate of step 1.2. The construction uses Riesz representation in $H_{0}^{1}$ only; no Hahn--Banach extension is invoked.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Uniformly elliptic divergence-form operators and their sesquilinear forms

Definition

Assume Countable Choice for the Sobolev interfaces. Let $Ω \subseteq R^{n}$ be open, $n \geq 1$ , and let $K \in {R, C}$ . Let $a^{i j}, b^{i}, c : Ω \to K$ , $i, j = 1, \dots, n$ , be measurable (A measurable function between measurable spaces) and essentially bounded (The space $L^{\infty} (μ)$ of essentially bounded measurable functions, The essential supremum of a measurable function with respect to a measure), with bounds $∣ a^{i j} ∣ \leq M_{a}, ∣ b^{i} ∣ \leq M_{b}, ∣ c ∣ \leq M_{c} a.e. on Ω,$ and suppose the uniform ellipticity condition holds: there is $θ > 0$ with $Re (\sum_{i, j = 1}^{n} a^{i j} (x) ξ_{j} \overline{ξ_{i}}) \geq θ ∣ ξ ∣^{2} for a.e. x \in Ω and all ξ \in C^{n} .$ The associated divergence-form expression is $L u : = - D_{i} (a^{i j} D_{j} u) + b^{i} D_{i} u + c u$ (Einstein summation over $i, j$ ), and the associated sesquilinear form on $H^{1} (Ω)$ is $a (u, v) : = \int_{Ω} (a^{i j} D_{j} u \overline{D_{i} v} + b^{i} D_{i} u \overline{v} + c u \overline{v}) d x .$ The form is linear in $u$ and conjugate-linear in $v$ , in the convention of Bounded, coercive and symmetric sesquilinear forms; the classical Dirichlet problem consists of $L u = f$ in $Ω$ with prescribed boundary values. The operator is determined by the coefficient functions only a.e., and all later statements about $a$ are statements about those classes. Where the domain and boundary data demand it (Weak Dirichlet solutions for a divergence-form operator, The inhomogeneous weak Dirichlet problem by a trace lifting) the domain is additionally a bounded $C^{1}$ domain (Bounded C^k domains and boundary charts). The convention is the complex sesquilinear one with conjugation in the second slot, as the plan's convention audit directs; the real case is the same with conjugation read as the identity.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Weak Dirichlet solutions for a divergence-form operator

Definition

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and let $L$ and its form $a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms on an open set $Ω \subseteq R^{n}$ . Homogeneous problem. Given $F \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ), a weak solution of $L u = F$ with zero boundary values is a class $u \in H_{0}^{1} (Ω)$ (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms) with $a (u, v) = F (v) for every v \in H_{0}^{1} (Ω) .$ Inhomogeneous problem. Additionally assume the Axiom of Choice (The Axiom of Choice) for the trace supplier, and let $n \geq 2$ and $Ω$ be a bounded $C^{1}$ domain (Bounded C^k domains and boundary charts), let $g \in H^{1 / 2} (\partial Ω) : = W^{1 / 2, 2} (\partial Ω)$ in the boundary scale of The fractional Sobolev space on a compact $C^{1}$ boundary and let $T$ be the trace operator of The $L^{p}$ trace operator on a bounded $C^{1}$ domain. A weak solution with boundary data $g$ is a class $u \in H^{1} (Ω)$ with $T u = g$ and $a (u, v) = F (v)$ for every $v \in H_{0}^{1} (Ω)$ . The defining identities are identities between functionals on the $H_{0}^{1}$ classes, so they are independent of the chosen almost-everywhere representatives of $u$ , of the coefficients and of the data (Weak differentiation ignores null-set changes, Complex Lp classes and Euclidean test-function conventions); the boundary condition is imposed through the trace of The $L^{p}$ trace operator on a bounded $C^{1}$ domain, never by pointwise evaluation. The Dirichlet condition is imposed by $u \in H_{0}^{1} (Ω)$ in the homogeneous problem and by $T u = g$ in the inhomogeneous problem; testing against $v \in H_{0}^{1}$ expresses the weak equation and does not by itself impose boundary data; the integrals are the ones proved absolutely convergent in The elliptic form is well defined and bounded on $H^{1}$ .

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

The elliptic form is well defined and bounded on $H^{1}$

Statement

Assume Countable Choice. Let $L$ and $a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with coefficient bounds $M_{a}, M_{b}, M_{c}$ (measurability and essential boundedness only; uniform ellipticity is not needed for this lemma). Then every term of $a (u, v) = \int_{Ω} (a^{i j} D_{j} u \overline{D_{i} v} + b^{i} D_{i} u \overline{v} + c u \overline{v}) d x$ is absolutely convergent for $u, v \in H^{1} (Ω)$ , the value depends only on the $H^{1}$ classes, and $a$ is a bounded sesquilinear form on $H^{1} (Ω)$ with $∣ a (u, v) ∣ \leq (n M_{a} + n M_{b} + M_{c}) ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}} .$ The same bound holds for the restriction of $a$ to $H_{0}^{1} (Ω)$ . The listed coefficient exponents are the whole hypothesis: no extra integrability of products is assumed.

Facts & Assumptions

Given: Countable Choice; an open $Ω \subseteq R^{n}$ , $n \geq 1$ ; coefficients $a^{i j}, b^{i}, c : Ω \to K$ measurable and essentially bounded with $∣ a^{i j} ∣ \leq M_{a}$ , $∣ b^{i} ∣ \leq M_{b}$ , $∣ c ∣ \leq M_{c}$ almost everywhere; and classes $u, v \in H^{1} (Ω) = W^{1, 2} (Ω; K)$ with weak derivatives $D_{j} u, D_{i} v$ .

[F1]

Coefficient hypotheses: each coefficient is a measurable, essentially bounded class with the stated a.e. bounds; the divergence-form operator and its form are those of Uniformly elliptic divergence-form operators and their sesquilinear forms (The essential supremum of a measurable function with respect to a measure, The space $L^{\infty} (μ)$ of essentially bounded measurable functions, A measurable function between measurable spaces).

[F2]

Sobolev norms: for $w \in H^{1} (Ω)$ the classes $w$ and $D_{j} w$ are in $L^{2} (Ω)$ , $∥ w ∥_{L^{2}} \leq ∥ w ∥_{H^{1}}$ and $∥ D_{j} w ∥_{L^{2}} \leq ∥ w ∥_{H^{1}}$ , and $H^{1} (Ω)$ is the a.e. quotient with quotient $L^{2}$ norms (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

[F3]

Weak derivatives depend only on the Sobolev class, and products of measurable classes are measurable and change, as integrands, only on null sets when representatives change (Weak differentiation ignores null-set changes, Complex Lp classes and Euclidean test-function conventions).

[F4]

Estimates: for real measurable $f, g$ Hölder gives $\int ∣ f g ∣ \leq ∥ f ∥_{p} ∥ g ∥_{p^{'}}$ for conjugate exponents, and the general product inequality gives $f g \in L^{r}$ with $∥ f g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ when $1 / r = 1 / p + 1 / q$ ; the complex forms are the componentwise ones of Complex Holder, Minkowski, and the quotient norm; for real vectors Cauchy--Schwarz gives $\sum_{j = 1}^{n} ∣ x_{j} ∣ \leq \sqrt{n} (\sum_{j = 1}^{n} ∣ x_{j} ∣^{2})^{1 / 2}$ (Holder's inequality for integrals, including the endpoint cases, Generalized Holder inequality puts products into $L^{r}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

1.1F1F3

Every integrand is measurable and bounded a.e. by a product of $L^{2}$ classes: the products $a^{i j} D_{j} u \overline{D_{i} v}$ , $b^{i} D_{i} u \overline{v}$ and $c u \overline{v}$ are measurable by [F1] and [F3], since products of measurable functions are measurable and representatives agree a.e.; the a.e. coefficient bounds turn each of them into an a.e. dominated multiple of a product of two $L^{2}$ classes, e.g. $∣ a^{i j} D_{j} u \overline{D_{i} v} ∣ \leq M_{a} ∣ D_{j} u ∣ ∣ D_{i} v ∣$ off a null set.

1.2F1F2F4

Principal part: for almost every $x$ , Cauchy--Schwarz in $R^{n}$ applied to the vectors $(∣ D_{j} u (x) ∣)_{j}$ and $(1, \dots, 1)$ , together with $∣ a^{i j} (x) ∣ \leq M_{a}$ , gives $∣ a^{i j} (x) D_{j} u (x) \overline{D_{i} v (x)} ∣ \leq n M_{a} ∣ D u (x) ∣ ∣ D v (x) ∣,$ where $∣ D u ∣ = (\sum_{j} ∣ D_{j} u ∣^{2})^{1 / 2}$ . Hence $\int_{Ω} ∣ a^{i j} D_{j} u \overline{D_{i} v} ∣ d x \leq n M_{a} ∥ D u ∥_{L^{2}} ∥ D v ∥_{L^{2}}$ by H"older with exponent $2$ , so the principal term converges absolutely.

1.3F1F2F4

Drift part: summing the coefficientwise bounds and applying H"older to each $∣ D_{i} u ∣ ∣ v ∣$ gives $\sum_{i = 1}^{n} \int_{Ω} ∣ b^{i} D_{i} u \overline{v} ∣ d x \leq M_{b} \sum_{i = 1}^{n} ∥ D_{i} u ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq n M_{b} ∥ D u ∥_{L^{2}} ∥ v ∥_{L^{2}}$ , since $∥ D_{i} u ∥_{L^{2}} \leq ∥ D u ∥_{L^{2}}$ ; the drift term is absolutely convergent.

1.4F1F2F4

Reaction part: $\int_{Ω} ∣ c u \overline{v} ∣ d x \leq M_{c} ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}}$ by H"older.

2.1F2step 1.2step 1.3step 1.4algebra

Bound: adding the three estimates and using $∥ D w ∥_{L^{2}} \leq ∥ w ∥_{H^{1}}$ and $∥ w ∥_{L^{2}} \leq ∥ w ∥_{H^{1}}$ for $w = u$ and $w = v$ gives $∣ a (u, v) ∣ \leq n M_{a} ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}} + n M_{b} ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}} + M_{c} ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}} = (n M_{a} + n M_{b} + M_{c}) ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ , so $a$ is a bounded form on $H^{1} (Ω)$ .

3.1F3step 2.1∎

Class independence and sesquilinearity: replacing $u$ , $v$ or any coefficient by another representative alters each integrand only on a null set, hence leaves every integral unchanged; in particular the two $L^{2}$ and weak-derivative slots depend only on the classes, and the value is finite by steps 1.2--2.1. Linearity in $u$ and conjugate-linearity in $v$ hold termwise: the principal and drift terms are linear in the $u$ -slot and conjugate-linear in the $v$ -slot, and the reaction term is linear in $u$ and conjugate-linear in $v$ , with the finite sum of absolutely convergent integrals linear separately in each slot. So $a$ is a well-defined bounded sesquilinear form on $H^{1} (Ω)$ ; on the subspace $H_{0}^{1} (Ω)$ the same estimate holds with the restricted norm.

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

Coercivity of the principal Dirichlet form

Statement

Assume the Axiom of Choice, inherited through the Poincaré supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open and bounded in one direction, and let $a_{0} (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ be the principal part of a uniformly elliptic form with constants $θ$ and $M_{a}$ (Uniformly elliptic divergence-form operators and their sesquilinear forms), restricted to $u, v \in H_{0}^{1} (Ω)$ . Then $a_{0}$ is a bounded sesquilinear form on $H_{0}^{1} (Ω)$ and $Re a_{0} (u, u) = \int_{Ω} Re (a^{i j} D_{j} u \overline{D_{i} u}) d x \geq θ ∥ D u ∥_{L^{2}}^{2} \geq \frac{θ}{1 + C_{P}^{2}} ∥ u ∥_{H_{0}^{1}}^{2} (u \in H_{0}^{1} (Ω)),$ where $C_{P}$ is the Poincar'e constant of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction for $p = 2$ . Hence $a_{0}$ is coercive on $H_{0}^{1} (Ω)$ with constant $α = θ / (1 + C_{P}^{2})$ , and $Re a_{0} (u, u) = θ ∥ D u ∥^{2}$ in the model case $a^{i j} = δ^{i j}$ (so $θ = 1$ ).

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open $Ω \subseteq R^{n}$ bounded in one direction; uniformly elliptic coefficients $a^{i j}$ with constants $θ > 0$ and $M_{a}$ (Uniformly elliptic divergence-form operators and their sesquilinear forms); and the principal form $a_{0} (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ on $H_{0}^{1} (Ω)$ .

[F1]

Uniform ellipticity: for almost every $x \in Ω$ and every $ξ \in C^{n}$ , $Re (\sum_{i, j} a^{i j} (x) ξ_{j} \overline{ξ_{i}}) \geq θ ∣ ξ ∣^{2}$ , and $∣ a^{i j} ∣ \leq M_{a}$ a.e. (Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F2]

Absolute convergence and boundedness: the principal term is absolutely convergent for $u, v \in H^{1} (Ω)$ and $∣ a_{0} (u, v) ∣ \leq n M_{a} ∥ u ∥_{H_{0}^{1}} ∥ v ∥_{H_{0}^{1}}$ on $H_{0}^{1} (Ω)$ ; in particular $a_{0}$ is a bounded sesquilinear form (The elliptic form is well defined and bounded on $H^{1}$ ).

[F3]

Poincar'e at $p = 2$ : $∥ u ∥_{L^{2} (Ω)} \leq C_{P} ∥ D u ∥_{L^{2} (Ω)}$ for every $u \in H_{0}^{1} (Ω)$ , where $D u = (D_{1} u, \dots, D_{n} u)$ and $C_{P}$ is the constant of the cited theorem at $p = 2$ ; hence $∥ u ∥_{H_{0}^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F4]

Nonnegative measurable functions have nonnegative integrals, and the integral of a nonnegative function is monotone under pointwise comparison; the real part of an integral of a complex-valued integrable function is the integral of its real part (Monotone convergence for the integral, Integral over a measurable subset, Real and imaginary parts, complex conjugation, and modulus).

[F5]

Coercivity of a sesquilinear form means $Re a (u, u) \geq α ∥ u ∥^{2}$ for all $u$ and some $α > 0$ (Bounded, coercive and symmetric sesquilinear forms).

Proof

1.1F1given

Pointwise bound: substituting $ξ = D u (x)$ in the ellipticity condition of [F1] and taking real parts gives, for almost every $x \in Ω$ , $Re (a^{i j} (x) D_{j} u (x) \overline{D_{i} u (x)}) \geq θ ∣ D u (x) ∣^{2} \geq 0.$

1.2F3

Poincar'e bound: by [F3], $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ for every $u \in H_{0}^{1} (Ω)$ , that is $∥ D u ∥_{L^{2}}^{2} \geq ∥ u ∥_{H_{0}^{1}}^{2} / (1 + C_{P}^{2})$ .

2.1F1F2F4step 1.1

Integrating the pointwise bound: the function $x \mapsto Re (a^{i j} D_{j} u \overline{D_{i} u})$ is measurable and its negative part is bounded by $(n M_{a} + n M_{a}) ∣ D u ∣^{2}$ a.e., so $Re a_{0} (u, u) = \int_{Ω} Re (a^{i j} D_{j} u \overline{D_{i} u}) d x$ ; the difference from $θ ∣ D u ∣^{2}$ is nonnegative and measurable, so its integral is nonnegative and $Re a_{0} (u, u) \geq θ \int_{Ω} ∣ D u ∣^{2} d x = θ ∥ D u ∥_{L^{2}}^{2}$ .

3.1F2F5step 2.1step 1.2algebra∎

Coercivity and the model case: combining steps 2.1 and 1.2 gives $Re a_{0} (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} \geq \frac{θ}{1 + C_{P}^{2}} ∥ u ∥_{H_{0}^{1}}^{2}$ for every $u \in H_{0}^{1} (Ω)$ , so $a_{0}$ is coercive with constant $α = θ / (1 + C_{P}^{2})$ ; it is bounded by [F2]. In the model case $a^{i j} = δ^{i j}$ , $θ = 1$ , the pointwise identity $Re (a^{i j} D_{j} u \overline{D_{i} u}) = ∣ D u ∣^{2}$ gives $Re a_{0} (u, u) = ∥ D u ∥_{L^{2}}^{2}$ .

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Testing a coercive weak solution with itself gives the energy bound

Statement

Let $H$ be a real or complex Hilbert space, $a$ a bounded coercive sesquilinear form with constant $α > 0$ , $F$ a bounded conjugate-linear functional on $H$ , and $u \in H$ a solution of $a (u, v) = F (v)$ for all $v \in H$ . Then $α ∥ u ∥^{2} \leq Re a (u, u) = Re F (u) \leq ∥ F ∥ ∥ u ∥, hence α ∥ u ∥ \leq ∥ F ∥ .$ The bound is a priori in the sense that it uses only the equation, coercivity and the norm of the datum, not the construction of $u$ ; it applies directly to homogeneous Dirichlet solutions $u \in H_{0}^{1} (Ω)$ of Weak Dirichlet solutions for a divergence-form operator after substituting their coercivity constants. For an inhomogeneous Dirichlet solution, first subtract a lifting to obtain a solution in $H_{0}^{1} (Ω)$ and use its residual datum; the original solution need not itself be an admissible test.

Facts & Assumptions

Given: A real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ with coercivity constant $α > 0$ ; a bounded conjugate-linear functional $F$ with $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ ; and a vector $u \in H$ with $a (u, v) = F (v)$ for every $v \in H$ .

[F1]

Coercivity: $Re a (u, u) \geq α ∥ u ∥^{2}$ ; and $Re F (u) \leq ∣ F (u) ∣ \leq ∥ F ∥ ∥ u ∥$ , since $Re z \leq ∣ z ∣$ (Bounded, coercive and symmetric sesquilinear forms, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Real and imaginary parts, complex conjugation, and modulus, A bounded linear operator between normed spaces, Hilbert space).

[F2]

The equation with the test $v = u$ reads $a (u, u) = F (u)$ (The Lax--Milgram theorem gives existence and uniqueness if Countable Choice is additionally assumed; here the identity uses only the assumed equation).

Proof

1.1F2given

Testing with the solution: substitute $v = u$ in the assumed equation, obtaining $a (u, u) = F (u)$ and hence, taking real parts, $Re a (u, u) = Re F (u)$ .

2.1F1step 1.1algebra∎

Two-sided bound: by coercivity, $α ∥ u ∥^{2} \leq Re a (u, u) = Re F (u) \leq ∣ F (u) ∣ \leq ∥ F ∥ ∥ u ∥$ . If $u \neq 0$ , divide by $∥ u ∥$ to obtain $α ∥ u ∥ \leq ∥ F ∥$ ; if $u = 0$ , the same inequality holds trivially. The estimate uses only the equation, coercivity and the datum norm, not any construction of $u$ .

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Existence and uniqueness for the weak Dirichlet Poisson problem

Statement

Assume the Axiom of Choice, inherited through the Poincaré supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, nonempty and bounded in one direction, with Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction). For every $F \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ) there is a unique $u \in H_{0}^{1} (Ω)$ with $\int_{Ω} \nabla u \cdot \overline{\nabla v} d x = F (v) for every v \in H_{0}^{1} (Ω),$ that is, the unique weak solution of $- Δ u = F$ with zero boundary values; it satisfies $∥ u ∥_{H_{0}^{1}} \leq (1 + C_{P}^{2}) ∥ F ∥_{H^{- 1}}, and ∥ D u ∥_{L^{2}} \leq (1 + C_{P}^{2})^{1 / 2} ∥ F ∥_{H^{- 1}} .$ Here $- Δ$ is the constant-coefficient operator $L u = - Δ u$ of Uniformly elliptic divergence-form operators and their sesquilinear forms, and the solution is the Lax--Milgram solution for the form $a_{0} (u, v) = \int \nabla u \cdot \overline{\nabla v}$ .

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open, nonempty $Ω \subseteq R^{n}$ bounded in one direction, with Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ ; the form $a_{0} (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H_{0}^{1} (Ω)$ ; and a functional $F \in H^{- 1} (Ω)$ , i.e. a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ .

[F1]

$H_{0}^{1} (Ω)$ is a Hilbert space for the $W^{1, 2}$ inner product, and $∥ u ∥_{H_{0}^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2}$ (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F2]

$a_{0}$ is the principal form with $a^{i j} = δ^{i j}$ , hence bounded on $H_{0}^{1}$ with bound $M = 1$ and coercive with constant $α = 1 / (1 + C_{P}^{2})$ (The elliptic form is well defined and bounded on $H^{1}$ , Coercivity of the principal Dirichlet form, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

Poincar'e: $∥ u ∥_{L^{2}} \leq C_{P} ∥ D u ∥_{L^{2}}$ on $H_{0}^{1}$ , so $∥ u ∥_{H_{0}^{1}} \leq (1 + C_{P}^{2})^{1 / 2} ∥ D u ∥_{L^{2}}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F4]

Lax--Milgram: bounded coercive forms on a Hilbert space with a bounded conjugate-linear datum have a unique solution, with $α ∥ u ∥ \leq ∥ F ∥$ for the coercivity constant $α$ (The Lax--Milgram theorem, The negative Sobolev space $H^{- 1} (Ω)$ , Weak Dirichlet solutions for a divergence-form operator).

[F5]

Energy identity for any solution: testing with itself gives $Re a_{0} (u, u) = Re F (u)$ and $α ∥ u ∥_{H_{0}^{1}}^{2} \leq Re a_{0} (u, u)$ , so $α ∥ u ∥_{H_{0}^{1}} \leq ∥ F ∥$ (Testing a coercive weak solution with itself gives the energy bound, Complex Lp classes and Euclidean test-function conventions).

Proof

1.1F1F2F4

Existence and uniqueness: by [F2] the form $a_{0}$ is bounded and coercive on the Hilbert space $H_{0}^{1} (Ω)$ ; applying Lax--Milgram [F4] to the bounded conjugate-linear functional $F \in H^{- 1} (Ω)$ gives a unique $u \in H_{0}^{1} (Ω)$ with $a_{0} (u, v) = F (v)$ for every $v \in H_{0}^{1} (Ω)$ , that is, the weak solution of $- Δ u = F$ in the sense of the definition.

1.2F2F5algebra

First estimate: by [F5] applied with $α = 1 / (1 + C_{P}^{2})$ , $∥ u ∥_{H_{0}^{1}} \leq (1 + C_{P}^{2}) ∥ F ∥_{H^{- 1}}$ .

2.1F3F5step 1.2algebra

Gradient estimate: the same substitution read as an identity gives $∥ D u ∥_{L^{2}}^{2} = Re a_{0} (u, u) = Re F (u) \leq ∥ F ∥ ∥ u ∥_{H_{0}^{1}} \leq (1 + C_{P}^{2})^{1 / 2} ∥ F ∥ ∥ D u ∥_{L^{2}}$ by Poincar'e [F3]; dividing by $∥ D u ∥_{L^{2}}$ when it is nonzero (and trivially otherwise) gives $∥ D u ∥_{L^{2}} \leq (1 + C_{P}^{2})^{1 / 2} ∥ F ∥_{H^{- 1}}$ .

3.1step 1.1step 1.2step 2.1∎

Conclusion: for every $F \in H^{- 1} (Ω)$ there is a unique weak solution of the zero-boundary Poisson problem, with the two displayed bounds; the solution is the Lax--Milgram solution for $a_{0}$ .

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Lax--Milgram solvability for coercive divergence-form equations

Statement

Assume the Axiom of Choice, inherited through the Poincaré supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, nonempty and bounded in one direction, let $L$ and $a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with ellipticity constant $θ$ , coefficient bounds $M_{a}, M_{b}, M_{c}$ (with the componentwise drift bounds $∣ b^{i} ∣ \leq M_{b}$ ), and let $C_{P}$ be the Poincar'e constant of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction for $p = 2$ . Assume the explicit smallness condition $θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0.$ Then for every $F \in H^{- 1} (Ω)$ there is a unique weak solution $u \in H_{0}^{1} (Ω)$ of $L u = F$ (Weak Dirichlet solutions for a divergence-form operator), and with $α_{0} : = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c}$ it satisfies $∥ u ∥_{H_{0}^{1}} \leq \frac{1 + C_{P}^{2}}{α_{0}} ∥ F ∥_{H^{- 1}} .$ When $b \equiv 0$ , taking $M_{b} = 0$ , the condition reduces to $M_{c} < θ / C_{P}^{2}$ , the sign/smallness condition of the plan; in the model case $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ , taking $θ = 1$ and $M_{b} = M_{c} = 0$ , it gives $α_{0} = 1$ and gives existence and uniqueness for the zero-boundary weak Poisson problem.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open, nonempty $Ω \subseteq R^{n}$ bounded in one direction; divergence-form coefficients with ellipticity constant $θ > 0$ and bounds $M_{a}, M_{b}, M_{c}$ , where $∣ b^{i} ∣ \leq M_{b}$ componentwise; the Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ at $p = 2$ ; the smallness assumption $α_{0} : = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0$ ; and the form $a$ on $H_{0}^{1} (Ω)$ .

[F1]

Pointwise ellipticity and coefficient bounds: $Re (a^{i j} D_{j} u \overline{D_{i} u}) \geq θ ∣ D u ∣^{2}$ a.e. and $∣ b^{i} ∣ \leq M_{b}$ , $∣ c ∣ \leq M_{c}$ a.e. (Uniformly elliptic divergence-form operators and their sesquilinear forms, The essential supremum of a measurable function with respect to a measure, The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[F2]

The form $a$ is bounded on $H_{0}^{1} (Ω)$ (The elliptic form is well defined and bounded on $H^{1}$ ) and $H_{0}^{1} (Ω)$ is a Hilbert space with $∥ u ∥_{H_{0}^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2}$ (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

Poincar'e: $∥ u ∥_{L^{2}} \leq C_{P} ∥ D u ∥_{L^{2}}$ for $u \in H_{0}^{1} (Ω)$ , hence $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F4]

Lax--Milgram and the a priori estimate: a bounded coercive form on a Hilbert space with a bounded conjugate-linear datum has a unique solution; any solution satisfies $α ∥ u ∥_{H_{0}^{1}} \leq ∥ F ∥$ for a coercivity constant $α$ (The Lax--Milgram theorem, Testing a coercive weak solution with itself gives the energy bound, The negative Sobolev space $H^{- 1} (Ω)$ , Weak Dirichlet solutions for a divergence-form operator).

[F5]

For $u \in H^{1} (Ω)$ , $\sum_{i = 1}^{n} ∥ D_{i} u ∥_{L^{2}} \leq \sqrt{n} ∥ D u ∥_{L^{2}}$ by Cauchy--Schwarz in the finite coordinate index (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

1.1F1F2F3F5algebra

Coercivity: for $u \in H_{0}^{1} (Ω)$ , pointwise ellipticity and the coefficient bounds give $Re a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} - \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}} - M_{c} ∥ u ∥_{L^{2}}^{2},$ since $∣ b^{i} ∣ \leq M_{b}$ and [F5] bound the coordinate sum. Poincar'e gives $∥ u ∥_{L^{2}} \leq C_{P} ∥ D u ∥_{L^{2}}$ , so the last two terms are at least $- \sqrt{n} C_{P} M_{b} ∥ D u ∥_{L^{2}}^{2}$ and $- C_{P}^{2} M_{c} ∥ D u ∥_{L^{2}}^{2}$ ; hence $Re a (u, u) \geq α_{0} ∥ D u ∥_{L^{2}}^{2}$ with $α_{0} = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c} > 0$ . Since $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ D u ∥_{L^{2}}^{2}$ , this gives $Re a (u, u) \geq \frac{α_{0}}{1 + C_{P}^{2}} ∥ u ∥_{H_{0}^{1}}^{2}$ : the form is coercive on $H_{0}^{1} (Ω)$ with constant $α_{0} / (1 + C_{P}^{2})$ , and it is bounded by [F2].

2.1F2F4step 1.1

Solvability: applying Lax--Milgram [F4] to the Hilbert space $H_{0}^{1} (Ω)$ , the bounded coercive form $a$ and the datum $F \in H^{- 1} (Ω)$ gives a unique $u \in H_{0}^{1} (Ω)$ with $a (u, v) = F (v)$ for every $v \in H_{0}^{1} (Ω)$ : a unique weak solution of $L u = F$ .

3.1F4step 1.1algebra∎

Estimate: the a priori estimate of [F4] with $α = α_{0} / (1 + C_{P}^{2})$ gives $∥ u ∥_{H_{0}^{1}} \leq \frac{1 + C_{P}^{2}}{α_{0}} ∥ F ∥_{H^{- 1}}$ . When $b \equiv 0$ , taking $M_{b} = 0$ , the condition is $M_{c} < θ / C_{P}^{2}$ , and in the model case $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ , taking $θ = 1$ and $M_{b} = M_{c} = 0$ , one has $α_{0} = 1$ , recovering the zero-boundary Poisson theorem.

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The Sobolev space $H^{1}$ is a Hilbert space

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, $n \geq 1$ , and $K \in {R, C}$ . On $H^{1} (Ω) = W^{1, 2} (Ω; K)$ (Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol) define $(u, v)_{H^{1}} : = (u, v)_{L^{2}} + \sum_{i = 1}^{n} (D_{i} u, D_{i} v)_{L^{2}},$ with the $L^{2}$ inner product of $L^{2}$ with the integral pairing is a Hilbert space. Then $(\cdot, \cdot)_{H^{1}}$ is an inner product on the Sobolev classes whose induced norm is the $W^{1, 2}$ norm of Integer-order Sobolev spaces and their norms, and $H^{1} (Ω)$ is a Hilbert space for it. The zero-boundary space $H_{0}^{1} (Ω)$ is a closed subspace of $H^{1} (Ω)$ (Zero-boundary Sobolev space as a norm closure) and hence a Hilbert space for the restricted inner product. The pairing is linear in the first argument and conjugate-linear in the second, in the convention of Real and complex inner-product spaces and their induced length.

Facts & Assumptions

Given: Countable Choice; an open $Ω \subseteq R^{n}$ , $n \geq 1$ ; a field $K \in {R, C}$ ; the space $H^{1} (Ω) = W^{1, 2} (Ω; K)$ with index set $A_{1} = {α \in N_{0}^{n} : ∣ α ∣ \leq 1} = {0, e_{1}, \dots, e_{n}}$ and the pairing $(u, v)_{H^{1}} : = (u, v)_{L^{2}} + \sum_{i = 1}^{n} (D_{i} u, D_{i} v)_{L^{2}}$ .

[F1]

Sobolev structure: each $D^{α} u$ is a well-defined $L^{2}$ class and the $W^{1, 2}$ norm is $∥ u ∥_{W^{1, 2}} = (\sum_{α \in A_{1}} ∥ D^{α} u ∥_{L^{2}}^{2})^{1 / 2}$ ; $H^{1} (Ω) = W^{1, 2} (Ω; K)$ by notation (Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol, The Sobolev norm descends to equivalence classes, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

$L^{2}$ is a Hilbert space for the integral pairing: on real $L^{2}$ the pairing $\int f g$ and on complex $L^{2}$ the pairing $\int f \overline{g}$ are well-defined inner products with $⟨ f, f ⟩ = ∥ f ∥_{2}^{2}$ , complete for the quotient $L^{2}$ norm ( $L^{2}$ with the integral pairing is a Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Complex Lp classes and Euclidean test-function conventions).

[F3]

An inner product is linear in the first argument, conjugate-symmetric, and positive definite; its induced length is a norm, and a Hilbert space is an inner-product space complete for that norm (Real and complex inner product spaces, with the inner product linear in the first argument, Real and complex inner-product spaces and their induced length, The induced length is a norm, Hilbert space).

[F4]

H"older: for $L^{2}$ classes $f, g$ , $∣ \int f \overline{g} ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ , with the real form $∣ \int f g ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ (Complex Holder, Minkowski, and the quotient norm, Holder's inequality for integrals, including the endpoint cases).

[F5]

$C_{c}^{\infty} (Ω; K)$ is a $K$ -vector space of test functions, and $H_{0}^{1} (Ω)$ is its closure in $H^{1}$ : explicitly, $u \in H_{0}^{1} (Ω)$ if and only if for every $δ > 0$ there is a test function $φ$ with $∥ u - φ ∥_{W^{1, 2}} < δ$ . A closure is closed and is the smallest closed superset (Complex Lp classes and Euclidean test-function conventions, Zero-boundary Sobolev space as a norm closure, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[F6]

A closed linear subspace of a Banach space, with the restricted norm, is a Banach space (A closed subspace of a Banach space is Banach).

Proof

1.1F1F2F3

The pairing is a well-defined inner product: each summand $(D^{α} u, D^{α} v)_{L^{2}}$ is the $L^{2}$ pairing of the well-defined classes $D^{α} u$ and $D^{α} v$ , hence representative-independent; each is linear in the first argument and conjugate-linear in the second over $K$ , and conjugate-symmetric. A finite sum of maps with these properties again has them, so $(u, v)_{H^{1}}$ is well defined on classes, linear in $u$ , conjugate-linear in $v$ and conjugate-symmetric. It is positive definite, since $(u, u)_{H^{1}} = \sum_{α \in A_{1}} ∥ D^{α} u ∥_{L^{2}}^{2} \geq 0$ equals $0$ only when every $D^{α} u = 0$ , in particular $u = D^{0} u = 0$ , while $u = 0$ plainly gives $0$ .

1.2F5algebra

$H_{0}^{1} (Ω)$ is a linear subspace. It contains the zero class, as the zero test function shows. Let $u, v \in H_{0}^{1} (Ω)$ and $a, b \in K$ , and let $δ > 0$ . Using the test-function approximation of [F5], choose test functions $φ, ψ$ with $∥ u - φ ∥_{W^{1, 2}} < δ / (2 (∣ a ∣ + ∣ b ∣ + 1))$ and $∥ v - ψ ∥_{W^{1, 2}} < δ / (2 (∣ a ∣ + ∣ b ∣ + 1))$ ; then $a φ + b ψ$ is again a test function, and $∥ (a u + b v) - (a φ + b ψ) ∥_{W^{1, 2}} \leq ∣ a ∣ ∥ u - φ ∥_{W^{1, 2}} + ∣ b ∣ ∥ v - ψ ∥_{W^{1, 2}} < δ$ . So $a u + b v \in H_{0}^{1} (Ω)$ ; the space is a subspace and, being a closure, it is closed in $H^{1} (Ω)$ .

2.1F1F2step 1.1algebra

Its induced norm is the Sobolev norm: $(u, u)_{H^{1}} = \sum_{α \in A_{1}} ∥ D^{α} u ∥_{L^{2}}^{2} = ∥ u ∥_{W^{1, 2}}^{2}$ , so the induced length is $∥ u ∥_{W^{1, 2}}$ .

3.1F1F2F4step 2.1

Completeness: let $(u_{m})$ be a Cauchy sequence in $H^{1}$ . For each $α \in A_{1}$ the inequality $∥ D^{α} u_{m} - D^{α} u_{l} ∥_{L^{2}} \leq ∥ u_{m} - u_{l} ∥_{W^{1, 2}}$ shows that $(D^{α} u_{m})_{m}$ is Cauchy in $L^{2}$ , so it has a limit class $f_{α}$ . Fix $i$ and a test function $φ$ . The weak-derivative identity gives $\int_{Ω} u_{m} D_{i} φ d x = - \int_{Ω} D_{i} u_{m} φ d x$ for every $m$ ; H"older's inequality makes both sides converge to $\int_{Ω} f_{0} D_{i} φ d x$ and $- \int_{Ω} f_{i} φ d x$ , respectively, where $f_{0}$ is the limit of the classes $u_{m} = D^{0} u_{m}$ . Hence $\int f_{0} D_{i} φ = - \int f_{i} φ$ for every test function $φ$ , so $f_{0} \in W^{1, 2} (Ω; K)$ with $D_{i} f_{0} = f_{i}$ , and $∥ u_{m} - f_{0} ∥_{W^{1, 2}}^{2} = \sum_{α \in A_{1}} ∥ D^{α} u_{m} - f_{α} ∥_{L^{2}}^{2} \to 0$ . Thus every Cauchy sequence in $H^{1}$ converges in $H^{1}$ : the space is complete in its Sobolev norm.

4.1F1F3step 1.1step 2.1step 3.1

Consequences for the pairing: by steps 1.1, 2.1 the pairing is an inner product inducing the Sobolev norm, and by step 3.1 the space is complete for that norm; therefore $H^{1} (Ω)$ is a Hilbert space over $K$ for the pairing.

5.1F3F5F6step 4.1step 1.2

$H_{0}^{1} (Ω)$ is a Hilbert space: it is a closed linear subspace of the Hilbert space $H^{1} (Ω)$ , hence complete for the restricted norm by [F6] applied to the underlying Banach space, and the restriction of the inner product is an inner product whose induced norm is the restriction of the Sobolev norm.

6.1step 4.1step 5.1∎

The pairing $(\cdot, \cdot)_{H^{1}}$ is therefore an inner product on $H^{1} (Ω)$ inducing the $W^{1, 2}$ norm, making $H^{1} (Ω)$ a Hilbert space, while $H_{0}^{1} (Ω)$ is a closed subspace and a Hilbert space for the restricted pairing; the pairing is linear in the first argument and conjugate-linear in the second, as required.

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Weak Neumann solvability on the mean-zero subspace

Statement

Assume the Axiom of Choice, inherited through the Poincaré supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ , $n \geq 1$ , be a nonempty bounded connected $W^{1, 2}$ -extension domain (Sobolev extension domains and extension operators), let $H^{1} (Ω)$ carry the inner product of The Sobolev space $H^{1}$ is a Hilbert space, and let $F$ be a bounded conjugate-linear functional on $H^{1} (Ω)$ with $F (1) = 0, 1 the constant function 1.$ Then there is a unique $u \in H^{1} (Ω)$ with $\int_{Ω} u d x = 0$ and $\int_{Ω} \nabla u \cdot \overline{\nabla v} d x = F (v) for every v \in H^{1} (Ω),$ the weak form of the homogeneous Neumann problem $- Δ u = F$ with $\partial_{ν} u = 0$ ; the full solution set is ${u + c 1 : c \in K}$ , and with $C_{W}$ the Poincar'e--Wirtinger constant of Poincare-Wirtinger on bounded connected extension domains by Rellich compactness, $∥ u ∥_{H^{1}} \leq (1 + C_{W}^{2}) ∥ F ∥ .$ The compatibility $F (1) = 0$ is necessary: constants lie in the kernel of the form, so if $F (1) \neq 0$ no solution exists. On a disconnected bounded $W^{1, 2}$ -extension domain there are finitely many connected components $Ω_{k}$ . Solvability is equivalent to $F (1_{Ω_{k}}) = 0$ for each component, with a unique solution having zero mean on each component; steps 1.3 and 4.2 prove this extension separately from the connected-domain Poincar'e--Wirtinger supplier.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded connected extension domain $Ω \subseteq R^{n}$ , $n \geq 1$ , with $Ω \neq \emptyset$ ; the Hilbert space $H^{1} (Ω)$ with inner product $(u, v)_{H^{1}} = (u, v)_{L^{2}} + \sum_{i} (D_{i} u, D_{i} v)_{L^{2}}$ ; the form $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ ; a bounded conjugate-linear functional $F$ on $H^{1} (Ω)$ with $F (1) = 0$ for the constant class $1$ ; and $V : = {v \in H^{1} (Ω) : \int_{Ω} v d x = 0}$ .

[F1]

$H^{1} (Ω)$ is a Hilbert space for the displayed inner product, whose induced norm is $∥ v ∥_{H^{1}}^{2} = ∥ v ∥_{L^{2}}^{2} + ∥ D v ∥_{L^{2}}^{2}$ with $∥ D v ∥_{L^{2}}^{2} = \sum_{i} ∥ D_{i} v ∥_{L^{2}}^{2}$ (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol).

[F2]

The linear functional $ℓ (v) = \int_{Ω} v d x$ is bounded on $H^{1} (Ω)$ : $∣ ℓ (v) ∣ \leq ∣ Ω ∣^{1 / 2} ∥ v ∥_{L^{2}} \leq ∣ Ω ∣^{1 / 2} ∥ v ∥_{H^{1}}$ by H"older, and $0 < ∣ Ω ∣ < \infty$ because $Ω$ is nonempty, open and bounded (Holder's inequality for integrals, including the endpoint cases, Euclidean balls have positive finite Lebesgue measure, Integral over a measurable subset).

[F3]

Poincar'e--Wirtinger with constant $C_{W} : = C (Ω, 2)$ : $∥ v ∥_{L^{2}} \leq C_{W} ∥ D v ∥_{L^{2}}$ for every $v \in V$ (Poincare-Wirtinger on bounded connected extension domains by Rellich compactness, Sobolev extension domains and extension operators).

[F4]

The constant class $1$ lies in $H^{1} (Ω)$ with weak gradient $0$ : its classical derivatives vanish and are its weak derivatives, and it is bounded on the finite-measure domain (Classical derivatives agree with weak derivatives, Complex Lp classes and Euclidean test-function conventions).

[F5]

Lax--Milgram: on a Hilbert space, a bounded coercive sesquilinear form with constant $α$ and a bounded conjugate-linear functional have a unique solution $u$ with $a (u, v) = F (v)$ for all $v$ , and $α ∥ u ∥ \leq ∥ F ∥$ (The Lax--Milgram theorem, A bounded linear operator between normed spaces, The dual space X^* of a normed space and its dual norm).

[F6]

Zero weak gradient implies componentwise constancy on each connected component; for the connected $Ω$ this says $\nabla w = 0$ a.e. implies $w$ is a constant class (Zero weak gradient gives componentwise constants).

[F7]

A closed linear subspace of a Hilbert space is a Hilbert space for the restricted inner product (A closed subspace of a Banach space is Banach, Linear subspace of a vector space, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent).

[F8]

On a bounded $W^{1, 2}$ -extension domain every $H^{1}$ -bounded sequence has an $L^{2}$ -convergent subsequence (Compactness of $W^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded extension domains). Components of an open Euclidean set are open, and their indicators are locally constant smooth functions with zero weak gradient (Every connected component of an open subset of $R^{n}$ is open and polygonally connected, Classical derivatives agree with weak derivatives).

Proof

1.1F2F7

$V$ is a closed subspace: $V = \ker ℓ$ for the bounded linear functional $ℓ$ of [F2], hence closed; being a linear subspace of the Hilbert space $H^{1} (Ω)$ , it is itself a Hilbert space for the restricted inner product by [F7].

1.2F1F3algebra

Coercivity on $V$ : for $v \in V$ , Poincar'e--Wirtinger gives $∥ v ∥_{H^{1}}^{2} = ∥ v ∥_{L^{2}}^{2} + ∥ D v ∥_{L^{2}}^{2} \leq (1 + C_{W}^{2}) ∥ D v ∥_{L^{2}}^{2}$ , so $Re a (v, v) = ∥ D v ∥_{L^{2}}^{2} \geq \frac{1}{1 + C_{W}^{2}} ∥ v ∥_{H^{1}}^{2}$ ; also $∣ a (u, v) ∣ \leq ∥ D u ∥_{L^{2}} ∥ D v ∥_{L^{2}} \leq ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ , so $a$ is bounded on $V$ with bound $1$ .

1.3F1F2F6F8given

Finiteness of components in the disconnected case. For a nonempty bounded $W^{1, 2}$ -extension domain, every component $C$ has positive measure by [F2] and its indicator belongs to $H^{1}$ with zero gradient by [F8]. If there were infinitely many components, AC would select distinct $C_{j}$ , $j \geq 1$ ; the normalized indicators $e_{j} = ∣ C_{j} ∣^{- 1 / 2} 1_{C_{j}}$ have $H^{1}$ norm $1$ and pairwise $L^{2}$ distance $\sqrt{2}$ , contradicting [F8]. Hence the components are $C_{1}, \dots, C_{m}$ . On the closed subspace $V_{c} = {v : \int_{C_{k}} v = 0 for every k}$ there is a constant $C_{c}$ with $∥ v ∥_{2} \leq C_{c} ∥ D v ∥_{2}$ . Otherwise AC selects $v_{j} \in V_{c}$ , $j \geq 1$ , with $∥ v_{j} ∥_{2} = 1$ and $∥ D v_{j} ∥_{2} < 1 / j$ . By [F8] a subsequence converges in $L^{2}$ ; it is Cauchy in $H^{1}$ , so [F1] gives an $H^{1}$ limit $v$ with $D v = 0$ and $∥ v ∥_{2} = 1$ . Each component integral passes to the limit by H"older, so $v \in V_{c}$ ; [F6] makes it constant on each component, hence zero by its componentwise mean, a contradiction.

2.1F5step 1.1step 1.2

Solution on $V$ : the restriction $F ∣_{V}$ is a bounded conjugate-linear functional on the Hilbert space $V$ and $a$ is bounded and coercive there with constant $α : = 1 / (1 + C_{W}^{2})$ ; Lax--Milgram gives a unique $u \in V$ with $a (u, v) = F (v)$ for every $v \in V$ , satisfying $α ∥ u ∥_{H^{1}} \leq ∥ F ∣_{V} ∥ \leq ∥ F ∥$ .

3.1F4step 2.1algebra

Extension to all test functions: let $v \in H^{1} (Ω)$ and put $c : = ∣ Ω ∣^{- 1} \int_{Ω} v d x$ and $v_{0} : = v - c 1$ , so that $\int v_{0} = 0$ and $v_{0} \in V$ , while $1 \in H^{1}$ has zero weak gradient by [F4]. Then $a (u, v) = a (u, v_{0}) + a (u, c 1) = a (u, v_{0})$ , and conjugate-linearity of $F$ together with $F (1) = 0$ gives $F (v) = F (v_{0}) + \overline{c} F (1) = F (v_{0})$ ; hence $a (u, v) = F (v)$ for every $v \in H^{1} (Ω)$ .

3.2step 1.2step 2.1

Uniqueness in $V$ : if $u_{1}, u_{2} \in V$ both solve, then $w : = u_{1} - u_{2} \in V$ satisfies $a (w, v) = 0$ for all $v \in V$ ; testing $v = w$ and using step 1.2 gives $α ∥ w ∥_{H^{1}}^{2} \leq Re a (w, w) = 0$ , so $w = 0$ .

3.3F5step 2.1algebra

Estimate: from step 2.1, $∥ u ∥_{H^{1}} \leq (1 + C_{W}^{2}) ∥ F ∣_{V} ∥ \leq (1 + C_{W}^{2}) ∥ F ∥$ , the last inequality because the supremum over the smaller set $V$ is at most the supremum over $H^{1} (Ω)$ .

4.1F1F6step 3.1algebra

Full solution set and necessity: if $u^{'}$ is any solution of $a (u^{'}, v) = F (v)$ on $H^{1} (Ω)$ , then $w : = u^{'} - u$ satisfies $a (w, v) = 0$ for all $v$ ; testing $v = w$ gives $∥ D w ∥_{L^{2}}^{2} = 0$ , so $\nabla w = 0$ a.e. and, $Ω$ being connected, [F6] makes $w$ a constant class; hence the solution set is $u + K 1$ , and conversely every $u + c 1$ solves because $1$ has zero weak gradient. Testing $v = 1$ in the equation gives $a (u, 1) = 0 = F (1)$ , so the compatibility $F (1) = 0$ is necessary.

4.2F5F6F8step 1.3step 1.1step 1.2step 2.1step 3.1algebra

Componentwise solvability. The inequality of step 1.3 gives coercivity on $V_{c}$ with constant $1 / (1 + C_{c}^{2})$ , so the argument of steps 1.1–2.1 gives a unique $u \in V_{c}$ solving there. Every $v \in H^{1}$ decomposes as $v = v_{0} + \sum_{k} c_{k} 1_{C_{k}}$ , where $c_{k} = ∣ C_{k} ∣^{- 1} \int_{C_{k}} v$ and $v_{0} \in V_{c}$ . Thus if $F (1_{C_{k}}) = 0$ for every $k$ , the equation extends to all tests as in step 3.1; conversely testing each indicator makes these conditions necessary. Testing the difference of two solutions with itself and using [F6] shows that all solutions differ by componentwise constants, so zero mean on each component specifies the unique normalized solution.

5.1step 4.1step 3.3step 1.3step 4.2∎

This proves the connected-domain assertion and its displayed estimate, and establishes the stated componentwise compatibility and normalization on disconnected bounded $W^{1, 2}$ -extension domains.

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A positive reaction term restores coercivity without Poincar'e

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open (no boundedness and no Dirichlet boundary condition assumed), and let $a$ be the divergence form of Uniformly elliptic divergence-form operators and their sesquilinear forms on $H^{1} (Ω)$ with ellipticity constant $θ$ , coefficient bounds $M_{a}, M_{b}, M_{c}$ where $∣ b^{i} ∣ \leq M_{b}$ componentwise, and ${ess inf}_{Ω} Re c \geq c_{0} > 0, \sqrt{n} M_{b} < 2 \min (θ, c_{0}) .$ Then $a$ is coercive on $Ω$ with constant $α : = \min (θ, c_{0}) - \sqrt{n} M_{b} / 2 > 0$ : $Re a (u, u) \geq α ∥ u ∥_{H^{1} (Ω)}^{2} (u \in H^{1} (Ω)) .$ Consequently The Lax--Milgram theorem applies on the Hilbert space $H^{1} (Ω)$ and gives, for every bounded conjugate-linear functional $F$ on $H^{1} (Ω)$ , a unique $u \in H^{1} (Ω)$ with $a (u, v) = F (v)$ for all $v$ : a second legitimate coercivity mechanism, driven by the reaction coefficient rather than by a Poincar'e inequality or boundary condition. When $b \equiv 0$ , taking $M_{b} = 0$ , the condition is $ess inf Re c > 0$ only.

Facts & Assumptions

Given: Countable Choice; an open $Ω \subseteq R^{n}$ ; divergence-form coefficients $a^{i j}, b^{i}, c$ with bounds $M_{a}, M_{b}, M_{c}$ , where $∣ b^{i} ∣ \leq M_{b}$ componentwise, and ellipticity constant $θ > 0$ ; ${ess inf}_{Ω} Re c \geq c_{0} > 0$ ; the assumption $\sqrt{n} M_{b} < 2 \min (θ, c_{0})$ ; and $α : = \min (θ, c_{0}) - \sqrt{n} M_{b} / 2 > 0$ .

[F1]

Uniform ellipticity gives for a.e. $x$ and $ξ = D u (x)$ : $Re (a^{i j} D_{j} u \overline{D_{i} u}) \geq θ ∣ D u ∣^{2}$ ; the coefficient bounds give $∣ b^{i} ∣ \leq M_{b}$ and $∣ c ∣ \leq M_{c}$ a.e. (Uniformly elliptic divergence-form operators and their sesquilinear forms, The essential supremum of a measurable function with respect to a measure, The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[F2]

The form $a$ is bounded on $H^{1} (Ω)$ by The elliptic form is well defined and bounded on $H^{1}$ , and $H^{1} (Ω)$ is a Hilbert space (The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms).

[F3]

Estimates: $Re \int_{Ω} c u \overline{u} d x = \int_{Ω} (Re c) ∣ u ∣^{2} d x \geq c_{0} ∥ u ∥_{L^{2}}^{2}$ ; componentwise $∣ b^{i} ∣ \leq M_{b}$ and $\sum_{i} ∣ D_{i} u ∣ \leq \sqrt{n} ∣ D u ∣$ imply $∣ \int_{Ω} b^{i} D_{i} u \overline{u} d x ∣ \leq \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}}$ by pointwise Cauchy--Schwarz and H"older; also $2 a b \leq a^{2} + b^{2}$ for nonnegative reals (Holder's inequality for integrals, including the endpoint cases, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, Real and imaginary parts, complex conjugation, and modulus).

[F4]

Lax--Milgram applies to bounded coercive forms on Hilbert spaces (The Lax--Milgram theorem, Bounded, coercive and symmetric sesquilinear forms).

Proof

1.1F1F2

Pointwise decomposition and integration: by [F1], for almost every $x$ the principal integrand satisfies $Re (a^{i j} D_{j} u \overline{D_{i} u}) \geq θ ∣ D u (x) ∣^{2}$ , and integrating (the principal term is absolutely convergent by [F2]) gives $Re \int_{Ω} a^{i j} D_{j} u \overline{D_{i} u} d x \geq θ ∥ D u ∥_{L^{2}}^{2} .$

2.1F1F3step 1.1algebra

Drift and reaction terms: taking real parts of the definition of $a$ , $Re a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} - \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}} + c_{0} ∥ u ∥_{L^{2}}^{2},$ where the drift term is bounded in absolute value by $\sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}}$ via [F3], and the reaction term is bounded below by $c_{0} ∥ u ∥_{L^{2}}^{2}$ using $Re c \geq c_{0}$ a.e.

3.1F3step 2.1algebra

Coercivity: applying $2 a b \leq a^{2} + b^{2}$ to $a = ∥ D u ∥_{L^{2}}$ , $b = ∥ u ∥_{L^{2}}$ with weight $\sqrt{n} M_{b}$ gives $\sqrt{n} M_{b} ∥ D u ∥ ∥ u ∥ \leq \frac{\sqrt{n} M_{b}}{2} (∥ D u ∥_{L^{2}}^{2} + ∥ u ∥_{L^{2}}^{2})$ , hence $Re a (u, u) \geq (θ - \frac{\sqrt{n} M_{b}}{2}) ∥ D u ∥_{L^{2}}^{2} + (c_{0} - \frac{\sqrt{n} M_{b}}{2}) ∥ u ∥_{L^{2}}^{2} \geq α ∥ u ∥_{H^{1} (Ω)}^{2}$ with $α = \min (θ, c_{0}) - \frac{\sqrt{n} M_{b}}{2} > 0$ , because $∥ u ∥_{H^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2}$ and both coefficients $θ - \frac{\sqrt{n} M_{b}}{2}$ , $c_{0} - \frac{\sqrt{n} M_{b}}{2}$ are at least $α$ by the smallness hypothesis.

4.1F2F4step 3.1∎

Consequences: $a$ is bounded by [F2] and coercive with constant $α$ by step 3.1, so Lax--Milgram applies on the Hilbert space $H^{1} (Ω)$ : for every bounded conjugate-linear functional $F$ there is a unique $u \in H^{1} (Ω)$ with $a (u, v) = F (v)$ for all $v$ . No Poincar'e inequality, boundary condition or integration by parts was used; when $b \equiv 0$ , taking $M_{b} = 0$ , the hypothesis reduces to $ess inf Re c > 0$ .

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The inhomogeneous weak Dirichlet problem by a trace lifting

Statement

Assume the Axiom of Choice (through the published Sobolev trace results) together with Countable Choice. Let $Ω \subset R^{n}$ , $n \geq 2$ , be a bounded $C^{1}$ domain, let $g \in H^{1 / 2} (\partial Ω) = W^{1 / 2, 2} (\partial Ω)$ and $F \in H^{- 1} (Ω)$ , and let $a$ be the divergence form of Uniformly elliptic divergence-form operators and their sesquilinear forms satisfying the coercivity condition of Lax--Milgram solvability for coercive divergence-form equations on $H_{0}^{1} (Ω)$ ; write $α : = α_{0} / (1 + C_{P}^{2}) > 0$ for its coercivity constant, where $α_{0} = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c}$ . Fix a bounded right inverse $R : H^{1 / 2} (\partial Ω) \to H^{1} (Ω)$ of the trace, $T \circ R = i d$ , as in A bounded right inverse of the trace, supported in a prescribed collar. Then there is a unique $u \in H^{1} (Ω)$ with $T u = g and a (u, v) = F (v) for every v \in H_{0}^{1} (Ω),$ and with $C_{a} = n M_{a} + n M_{b} + M_{c}$ the bound of The elliptic form is well defined and bounded on $H^{1}$ on all of $H^{1}$ , $∥ u ∥_{H^{1}} \leq ∥ R g ∥_{H^{1}} + \frac{∥ F ∥_{H^{- 1}} + C_{a} ∥ R g ∥_{H^{1}}}{α} \leq C (Ω, a, R) (∥ g ∥_{W^{1 / 2, 2}} + ∥ F ∥_{H^{- 1}}) .$ The solution is independent of the choice of lifting; the displayed estimate depends on the fixed right inverse $R$ . Boundary data outside the trace range $H^{1 / 2} (\partial Ω)$ are not admissible: no $H^{1}$ function has such a trace, the trace range being exactly $H^{1 / 2} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space).

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded $C^{1}$ domain $Ω \subset R^{n}$ , $n \geq 2$ ; boundary data $g \in H^{1 / 2} (\partial Ω) = W^{1 / 2, 2} (\partial Ω)$ ; $F \in H^{- 1} (Ω)$ ; a divergence form $a$ on $H^{1} (Ω)$ whose restriction to $H_{0}^{1} (Ω)$ satisfies the coercivity condition of Lax--Milgram solvability for coercive divergence-form equations with $α = α_{0} / (1 + C_{P}^{2}) > 0$ , $α_{0} = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c}$ , and bounded with constant $C_{a}$ ; and a bounded right inverse $R$ of the trace $T$ with $T \circ R = i d$ .

[F1]

The trace operator $T : H^{1} (Ω) \to H^{1 / 2} (\partial Ω)$ is bounded and its kernel is exactly $H_{0}^{1} (Ω)$ (The kernel of the trace is the closure of the test functions, The fractional Sobolev space on a compact $C^{1}$ boundary, Bounded C^k domains and boundary charts).

[F2]

The right inverse satisfies $T (R g) = g$ and $∥ R g ∥_{H^{1}} \leq ∥ R ∥ ∥ g ∥_{W^{1 / 2, 2}}$ (A bounded right inverse of the trace, supported in a prescribed collar, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

The divergence-form solvability theorem: for every datum in $H^{- 1} (Ω)$ there is a unique $w \in H_{0}^{1} (Ω)$ with $a (w, v) = G (v)$ for all $v \in H_{0}^{1} (Ω)$ , satisfying $∥ w ∥_{H_{0}^{1}} \leq ∥ G ∥_{H^{- 1}} / α$ (Lax--Milgram solvability for coercive divergence-form equations, Weak Dirichlet solutions for a divergence-form operator).

[F4]

Boundedness on all $H^{1}$ slots: with $C_{a} = n M_{a} + n M_{b} + M_{c}$ , $∣ a (u, v) ∣ \leq C_{a} ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ for every $u, v \in H^{1} (Ω)$ (The elliptic form is well defined and bounded on $H^{1}$ , The negative Sobolev space $H^{- 1} (Ω)$ , Complex Lp classes and Euclidean test-function conventions, Holder's inequality for integrals, including the endpoint cases).

[F5]

Admissibility is exactly trace-range membership: the trace operator $T : H^{1} (Ω) \to H^{1 / 2} (\partial Ω)$ has range exactly $H^{1 / 2} (\partial Ω)$ (The $L^{p}$ trace operator on a bounded $C^{1}$ domain, The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact $C^{1}$ boundary), so a datum outside $H^{1 / 2} (\partial Ω)$ is the trace of no $H^{1}$ function and the inhomogeneous problem admits no solution for it.

Proof

1.1F2F4algebra

Lift and shift: put $u_{0} : = R g$ , so $T u_{0} = g$ and $∥ u_{0} ∥_{H^{1}} \leq ∥ R ∥ ∥ g ∥$ ; define $\tilde{F} (v) : = F (v) - a (u_{0}, v)$ for $v \in H_{0}^{1} (Ω)$ . Then $\tilde{F}$ is conjugate-linear, and by [F4] $∥ \tilde{F} ∥_{H^{- 1}} \leq ∥ F ∥_{H^{- 1}} + C_{a} ∥ u_{0} ∥_{H^{1}} .$

2.1F3step 1.1

Zero-boundary correction: by [F3] applied to $\tilde{F}$ there is a unique $w \in H_{0}^{1} (Ω)$ with $a (w, v) = \tilde{F} (v)$ for all $v \in H_{0}^{1} (Ω)$ , and $∥ w ∥_{H^{1}} \leq ∥ \tilde{F} ∥_{H^{- 1}} / α$ .

3.1F1step 2.1algebra

The sum solves the inhomogeneous problem: let $u : = u_{0} + w$ . Since $w \in H_{0}^{1} (Ω) = \ker T$ by [F1], $T u = T u_{0} + T w = g$ . For $v \in H_{0}^{1} (Ω)$ , additivity of $a$ in the first slot gives $a (u, v) = a (u_{0}, v) + a (w, v) = a (u_{0}, v) + \tilde{F} (v) = F (v)$ .

3.2F2step 1.1step 2.1algebra

Estimate: $∥ u ∥_{H^{1}} \leq ∥ u_{0} ∥_{H^{1}} + ∥ w ∥_{H^{1}} \leq ∥ R g ∥_{H^{1}} + (∥ F ∥_{H^{- 1}} + C_{a} ∥ R g ∥_{H^{1}}) / α$ , and the right-inverse bound $∥ R g ∥_{H^{1}} \leq ∥ R ∥ ∥ g ∥_{W^{1 / 2, 2}}$ makes the right-hand side at most $C (Ω, a, R) (∥ g ∥_{W^{1 / 2, 2}} + ∥ F ∥_{H^{- 1}})$ for an explicit constant depending only on $Ω$ , $a$ and $R$ .

4.1F1step 3.1

Uniqueness independent of the lifting: if $u_{1}, u_{2}$ are solutions, then $z : = u_{1} - u_{2}$ has $T z = 0$ , so $z \in H_{0}^{1} (Ω)$ by [F1], and $a (z, v) = 0$ for every $v \in H_{0}^{1} (Ω)$ . Testing $v = z$ and using coercivity gives $α ∥ z ∥_{H^{1}}^{2} \leq Re a (z, z) = 0$ , so $z = 0$ .

5.1F5given∎

Admissibility: the construction needs $g$ in the trace range; by [F5] a datum outside $H^{1 / 2} (\partial Ω)$ is the trace of no $H^{1}$ function, so the inhomogeneous problem has no solution for it and the trace-range hypothesis cannot be dropped.

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Classical solutions satisfy the weak formulation

Statement

Assume the Axiom of Choice (through the published Sobolev Gauss--Green formula) and Countable Choice. Let $Ω \subset R^{n}$ , $n \geq 2$ , be a bounded $C^{1}$ domain, let $a^{i j} \in C^{1} (\overline{Ω})$ , $b^{i}, c \in C (\overline{Ω})$ with uniform ellipticity constant $θ$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, Bounded C^k domains and boundary charts), let $u \in C^{2} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ and $f \in C (\overline{Ω})$ , and set $L u : = - D_{i} (a^{i j} D_{j} u) + b^{i} D_{i} u + c u .$ If $L u = f$ on $Ω$ , then $u$ is a weak solution in the sense of Weak Dirichlet solutions for a divergence-form operator for the datum $F_{f} (v) : = (f, v)_{L^{2}}$ : $a (u, v) = \int_{Ω} f \overline{v} d x for every v \in H_{0}^{1} (Ω) .$ No converse is claimed: the lemma is the classical-to-weak consistency statement only.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded $C^{1}$ domain $Ω \subset R^{n}$ , $n \geq 2$ ; coefficients $a^{i j} \in C^{1} (\overline{Ω})$ , $b^{i}, c \in C (\overline{Ω})$ with ellipticity constant $θ$ ; a class $u \in C^{2} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ and $f \in C (\overline{Ω})$ with $L u : = - D_{i} (a^{i j} D_{j} u) + b^{i} D_{i} u + c u = f$ on $Ω$ ; the divergence form $a (u, v) = \int_{Ω} (a^{i j} D_{j} u \overline{D_{i} v} + b^{i} D_{i} u \overline{v} + c u \overline{v}) d x$ ; and the trace operator $T$ of The $L^{p}$ trace operator on a bounded $C^{1}$ domain with outward normal $ν$ .

[F1]

$u \in C^{2} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ has classical derivatives that are its weak derivatives, and likewise $a^{i j} D_{j} u \in C^{1} (\overline{Ω})$ has for each $i$ the classical derivative $D_{i} (a^{i j} D_{j} u) \in C (\overline{Ω})$ as its weak derivative (Classical derivatives agree with weak derivatives, $C^{k}$ maps and multi-index derivative notation in Euclidean space, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F2]

Kernel of the trace: for $1 \leq p < \infty$ the kernel of $T$ on $W^{1, p} (Ω)$ is exactly $W_{0}^{1, p} (Ω)$ ; in particular $w \in H_{0}^{1} (Ω)$ implies $T w = 0$ (The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure).

[F3]

$H_{0}^{1} (Ω)$ is stable under conjugation, being the closure of the conjugation-stable space $C_{c}^{\infty} (Ω; K)$ ; complex weak derivatives are taken componentwise, so $D_{i} \overline{w} = \overline{D_{i} w}$ for $w \in H^{1}$ (Zero-boundary Sobolev space as a norm closure, Complex Lp classes and Euclidean test-function conventions, Integer-order Sobolev spaces and their norms).

[F4]

Sobolev Gauss--Green: for $n \geq 2$ , $1 < p < \infty$ , $U \in W^{1, p} (Ω; K)$ and $W \in W^{1, p^{'}} (Ω; K)$ one has $\int_{Ω} U D_{i} W d x = - \int_{Ω} (D_{i} U) W d x + \int_{\partial Ω} (T U) (T W) ν_{i} d S$ , all integrals finite (The Gauss-Green integration-by-parts formula with Sobolev traces, Bounded C^k domains and boundary charts, Classical normal derivative).

[F5]

The datum $F_{f} (v) : = (f, v)_{L^{2}}$ is a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ , i.e. an element of $H^{- 1} (Ω)$ : $∣ F_{f} (v) ∣ \leq ∥ f ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq ∥ f ∥_{L^{2}} ∥ v ∥_{H_{0}^{1}}$ , and $Ω$ is bounded hence bounded in one direction ( $L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound, Holder's inequality for integrals, including the endpoint cases, Weak Dirichlet solutions for a divergence-form operator, The space $L^{p} (μ)$ as the quotient by null functions).

Proof

1.1F2F3

Boundary term vanishes: let $v \in H_{0}^{1} (Ω)$ . Then $\overline{v} \in H_{0}^{1} (Ω)$ by conjugation stability, so $T \overline{v} = 0$ ; consequently every boundary term carrying the factor $T \overline{v}$ vanishes.

2.1F1F3F4step 1.1

Gauss--Green for one coefficient: fix $i$ . Since $a^{i j} D_{j} u \in C^{1} (\overline{Ω}) \subseteq W^{1, 2} (Ω)$ and $\overline{v} \in W^{1, 2} (Ω)$ , applying the Gauss--Green formula with $U = a^{i j} D_{j} u$ and $W = \overline{v}$ gives $\int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x = - \int_{Ω} D_{i} (a^{i j} D_{j} u) \overline{v} d x + \int_{\partial Ω} T (a^{i j} D_{j} u) T (\overline{v}) ν_{i} d S,$ and the boundary integral is $0$ by step 1.1. Summing over $i, j$ and noting that $D_{i} \overline{v} = \overline{D_{i} v}$ and $D_{j} u$ are the weak derivatives of the classical ones yields $\int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x = - \int_{Ω} D_{i} (a^{i j} D_{j} u) \overline{v} d x$ .

3.1F1F3step 2.1algebra

Weak equation: adding the drift and reaction terms and using the classical derivatives as weak derivatives, $a (u, v) = \int_{Ω} (- D_{i} (a^{i j} D_{j} u) + b^{i} D_{i} u + c u) \overline{v} d x = \int_{Ω} (L u) \overline{v} d x = \int_{Ω} f \overline{v} d x$ for every $v \in H_{0}^{1} (Ω)$ ; here $L u = f$ holds as an identity of continuous functions on $Ω$ by hypothesis, and $F_{f} (v) = \int_{Ω} f \overline{v} d x$ is the $L^{2}$ pairing.

4.1F5step 3.1∎

Conclusion: by [F5] the functional $F_{f}$ lies in $H^{- 1} (Ω)$ , and step 3.1 exhibits $a (u, v) = F_{f} (v)$ for every $v \in H_{0}^{1} (Ω)$ with $u \in C^{2} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ ; hence every classical solution with $L u = f$ is a weak solution in the sense of the definition. No converse is claimed.

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Weak solutions depend continuously on the data

Statement

Assume the Axiom of Choice and Countable Choice, and take the hypotheses of The inhomogeneous weak Dirichlet problem by a trace lifting with one fixed right inverse $R$ and coercivity constant $α = α_{0} / (1 + C_{P}^{2})$ (the constant displayed there, where $α_{0} = θ - \sqrt{n} C_{P} M_{b} - C_{P}^{2} M_{c}$ ). If $u_{i}$ solve the weak Dirichlet problems with data $(F_{i}, g_{i}) \in H^{- 1} (Ω) \times H^{1 / 2} (\partial Ω)$ , $i = 1, 2$ , then $∥ u_{1} - u_{2} ∥_{H^{1}} \leq (1 + \frac{C_{a}}{α}) ∥ R (g_{1} - g_{2}) ∥_{H^{1}} + \frac{∥ F_{1} - F_{2} ∥_{H^{- 1}}}{α} \leq C (Ω, a) (∥ F_{1} - F_{2} ∥_{H^{- 1}} + ∥ g_{1} - g_{2} ∥_{W^{1 / 2, 2}}),$ where $C_{a} = n M_{a} + n M_{b} + M_{c}$ is the full $H^{1}$ bound of the form and $C (Ω, a)$ depends only on the domain, the coefficients, the coercivity constant and the fixed right inverse. Thus the solution map is Lipschitz on the product of the data spaces, and $u_{1} = u_{2}$ when the data agree.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded $C^{1}$ domain $Ω \subset R^{n}$ , $n \geq 2$ ; a divergence form $a$ bounded on $H^{1} (Ω)$ with constant $C_{a}$ and coercive on $H_{0}^{1} (Ω)$ with constant $α > 0$ ; a fixed bounded right inverse $R$ of the trace with $T \circ R = i d$ ; and solutions $u_{1}, u_{2} \in H^{1} (Ω)$ of the weak problems with data $(F_{i}, g_{i})$ .

[F1]

Each $u_{i}$ satisfies $T u_{i} = g_{i}$ and $a (u_{i}, v) = F_{i} (v)$ for every $v \in H_{0}^{1} (Ω)$ , and solve the problem via the lifting construction: $u_{i} = R g_{i} + w_{i}$ with $w_{i} \in H_{0}^{1} (Ω)$ (The inhomogeneous weak Dirichlet problem by a trace lifting, Weak Dirichlet solutions for a divergence-form operator).

[F2]

Linearity and boundedness: the same $C_{a}$ from The elliptic form is well defined and bounded on $H^{1}$ bounds $a$ on all of $H^{1} (Ω) \times H^{1} (Ω)$ , and hence also on the restriction to $H_{0}^{1} (Ω)$ . Thus $v \mapsto (F_{1} - F_{2}) (v) - a (R (g_{1} - g_{2}), v)$ is a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ of norm at most $∥ F_{1} - F_{2} ∥_{H^{- 1}} + C_{a} ∥ R (g_{1} - g_{2}) ∥_{H^{1}}$ (The negative Sobolev space $H^{- 1} (Ω)$ , The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F3]

A priori bound: any $w \in H_{0}^{1} (Ω)$ with $a (w, v) = \tilde{F} (v)$ for all $v \in H_{0}^{1} (Ω)$ satisfies $α ∥ w ∥_{H^{1}} \leq ∥ \tilde{F} ∥_{H^{- 1}}$ (Testing a coercive weak solution with itself gives the energy bound, Lax--Milgram solvability for coercive divergence-form equations).

[F4]

Right-inverse bound: $∥ R (g_{1} - g_{2}) ∥_{H^{1}} \leq ∥ R ∥ ∥ g_{1} - g_{2} ∥_{W^{1 / 2, 2}}$ (A bounded right inverse of the trace, supported in a prescribed collar, The fractional Sobolev space on a compact $C^{1}$ boundary, Integer-order Sobolev spaces and their norms).

[F5]

The abstract estimate is the operator-norm bound for the Lax--Milgram solution operator (The Lax--Milgram solution operator has norm at most $1 / α$ , The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

1.1F1F2algebra

The difference solves a shifted problem: put $g : = g_{1} - g_{2}$ , $F : = F_{1} - F_{2}$ and $w : = (u_{1} - u_{2}) - R g$ . Then $T w = T (u_{1} - u_{2}) - T R g = g_{1} - g_{2} - g = 0$ , so $w \in H_{0}^{1} (Ω)$ , and for every $v \in H_{0}^{1} (Ω)$ first-slot linearity gives $a (w, v) = a (u_{1}, v) - a (u_{2}, v) - a (R g, v) = F (v) - a (R g, v) = : \tilde{F} (v) .$

1.2F2

Bound on the shifted datum: by [F2] the functional $\tilde{F}$ is bounded and conjugate-linear with $∥ \tilde{F} ∥_{H^{- 1}} \leq ∥ F ∥_{H^{- 1}} + C_{a} ∥ R g ∥_{H^{1}}$ .

2.1F3F4step 1.2algebra

Energy estimate and conclusion: the a priori bound [F3] applied to $w$ and $\tilde{F}$ gives $∥ w ∥_{H^{1}} \leq ∥ \tilde{F} ∥_{H^{- 1}} / α$ , hence by the triangle inequality and [F4] $∥ u_{1} - u_{2} ∥_{H^{1}} \leq ∥ R g ∥_{H^{1}} + \frac{∥ F ∥_{H^{- 1}} + C_{a} ∥ R g ∥_{H^{1}}}{α} \leq (1 + \frac{C_{a}}{α}) ∥ R ∥ ∥ g_{1} - g_{2} ∥_{W^{1 / 2, 2}} + \frac{∥ F_{1} - F_{2} ∥_{H^{- 1}}}{α},$ which is the first display; the second follows by absorbing $1 + C_{a} / α$ and $∥ R ∥$ into the constant $C (Ω, a)$ . If the data agree then $g = 0$ , $F = 0$ , $\tilde{F} = 0$ and $w = 0$ , so $u_{1} = u_{2}$ .

3.1F5step 2.1∎

Operator form: the same estimate is the statement that the solution map $(F, g) \mapsto u$ is Lipschitz on $H^{- 1} (Ω) \times H^{1 / 2} (\partial Ω)$ with the displayed constant; the abstract mechanism is the norm bound $∥ S ∥ \leq 1 / α$ for the Lax--Milgram solution operator on the zero-boundary part.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The sharp Dirichlet Poincare inequality on an interval

Statement

Assume Countable Choice. Let $L > 0$ , $I = (0, L)$ and $K \in {R, C}$ . Then every $u \in H_{0}^{1} (I; K)$ satisfies $∥ u ∥_{L^{2} (I)} \leq \frac{L}{π} ∥ u^{'} ∥_{L^{2} (I)} .$ The constant $L / π$ is optimal: $ϕ (x) = \sin (π x / L)$ belongs to $H_{0}^{1} (I)$ , is nonzero, and attains equality. Moreover $\int_{I} ϕ^{'} \overline{v^{'}} d x = (π / L)^{2} \int_{I} ϕ \overline{v} d x (v \in H_{0}^{1} (I)) .$ This is a direct interval inequality and weak identity; no spectral decomposition is assumed.

Facts & Assumptions

Given: Reals $L > 0$ and $k : = π / L$ , the interval $I = (0, L)$ , a field $K \in {R, C}$ , and the function $ϕ (x) : = \sin (k x)$ .

[F1]

$H_{0}^{1} (I; K) = W_{0}^{1, 2} (I; K)$ is the closure of $C_{c}^{\infty} (I; K)$ in the $W^{1, 2}$ norm, which for functions of one variable is $∥ v ∥_{H^{1}}^{2} = ∥ v ∥_{L^{2} (I)}^{2} + ∥ v^{'} ∥_{L^{2} (I)}^{2}$ ; complex test functions are defined by requiring both components to be real test functions, and the $L^{2}$ theory of complex classes is the componentwise one (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, Complex Lp classes and Euclidean test-function conventions).

[F2]

A classical smooth derivative on $I$ is the weak derivative of the class (Classical derivatives agree with weak derivatives). On a closed bounded interval a bounded Riemann integrable function is Lebesgue measurable and Lebesgue integrable with the same integral, a continuous function on $[a, b]$ is Riemann integrable, the Riemann integral is linear and additive over subintervals, and $m \leq h \leq M$ on $[a, b]$ gives $m (b - a) \leq \int_{a}^{b} h \leq M (b - a)$ (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ).

[F3]

Integration by parts and the second fundamental theorem on $[a, b]$ for differentiable functions with integrable derivatives (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F4]

Trigonometric facts: $(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , $\sin 0 = 0$ ; $\sin π = 0$ and $\sin x > 0$ for $0 < x < π$ ; $\sin^{2} x + \cos^{2} x = 1$ , $∣ \sin x ∣, ∣ \cos x ∣ \leq 1$ ; $∣ \sin u - \sin v ∣ \leq ∣ u - v ∣$ ; and the addition formulas, in particular $\cos 2 x = \cos^{2} x - \sin^{2} x$ and $\sin 2 π = 2 \sin π \cos π$ (The derivatives of sine and cosine are cosine and minus sine, Pi is the first positive zero of sine, Parity and the Pythagorean identity for sine and cosine, Sine and cosine are $1$ -Lipschitz on $R$ , The addition formulas for sine and cosine).

[F5]

Chain rule and the algebra of derivatives (sums, products, quotients away from zeros) for real functions (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F6]

The standard smooth step $σ$ is smooth with $σ (t) = 0$ for $t \leq 0$ and $σ (t) = 1$ for $t \geq 1$ ; hence $σ^{'} \equiv 0$ outside $(0, 1)$ and, being continuous on the compact interval $[0, 1]$ , the derivative satisfies $C_{σ} : = \sup_{R} ∣ σ^{'} ∣ < \infty$ (The standard smooth step function, Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value).

[F7]

H"older's inequality and the quotient norms: for complex $L^{2}$ classes, $∣ \int f \overline{g} ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ , and the quotient norm obeys the triangle inequality, hence $∣ ∥ f ∥_{2} - ∥ g ∥_{2} ∣ \leq ∥ f - g ∥_{2}$ (Complex Holder, Minkowski, and the quotient norm).

[F8]

Countable Choice is assumed (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1F4F5

Properties of $ϕ$ : differentiating $ϕ = \sin \circ (k i d)$ gives $ϕ^{'} (x) = k \cos (k x)$ and $ϕ^{''} (x) = - k^{2} \sin (k x) = - k^{2} ϕ (x)$ by the chain rule; $∣ ϕ^{'} ∣ \leq k$ because $∣ \cos ∣ \leq 1$ ; $ϕ > 0$ on $I$ because $0 < k x < π$ there; and the one-Lipschitz property of sine at the points $0$ , $k x$ and $π$ gives $∣ ϕ (x) ∣ = ∣ \sin (k x) - \sin 0 ∣ \leq k x$ and $∣ ϕ (x) ∣ = ∣ \sin (k x) - \sin π ∣ \leq k (L - x)$ for $x \in I$ .

1.2F2F3F4F5algebra

$ϕ$ is a nonzero class: $\sin^{2} t = \frac{1}{2} - \frac{1}{2} \cos 2 t$ and $\cos 2 t = 1 - 2 \sin^{2} t$ from the addition formulas and the Pythagorean identity give $ϕ^{2} = \frac{1}{2} - \frac{1}{2} \cos (2 k x)$ ; the antiderivative $x \mapsto \frac{\sin (2 k x)}{2 k}$ has derivative $\cos (2 k x)$ by the chain rule and vanishes at $x = 0$ and at $x = L$ , where $\sin (2 π) = 2 \sin π \cos π = 0$ . So $\int_{0}^{L} ϕ^{2} = \frac{L}{2} - \frac{1}{2} \int_{0}^{L} \cos (2 k x) d x = \frac{L}{2} > 0$ by linearity, the fundamental theorem and elementary bounds; in particular the $L^{2}$ class of $ϕ$ is not zero.

2.1F2F3F5step 1.1algebra

Real smooth case: let $a \in C_{c}^{\infty} (I; R)$ and put $w : = a / ϕ$ , a real $C_{c}^{\infty}$ function, since $supp a$ is a compact subset of $I$ on which $ϕ > 0$ . Then $a = ϕ w$ , so $a^{'} = ϕ^{'} w + ϕ w^{'}$ and $(∣ a^{'} ∣^{2} - k^{2} ∣ a ∣^{2}) = ϕ^{2} ∣ w^{'} ∣^{2} + (ϕ ϕ^{'} w^{2})^{'}$ : indeed $∣ a^{'} ∣^{2} = ϕ^{' 2} w^{2} + 2 ϕ ϕ^{'} w w^{'} + ϕ^{2} w^{' 2}$ while $(ϕ ϕ^{'} w^{2})^{'} = (ϕ^{' 2} + ϕ ϕ^{''}) w^{2} + 2 ϕ ϕ^{'} w w^{'} = (ϕ^{' 2} - k^{2} ϕ^{2}) w^{2} + 2 ϕ ϕ^{'} w w^{'}$ by step 1.1. Choose endpoints $r < s$ in $I$ with $supp a \subseteq (r, s)$ ; then $ϕ ϕ^{'} w^{2}$ vanishes at $r$ and $s$ , so the fundamental theorem gives $\int_{r}^{s} (ϕ ϕ^{'} w^{2})^{'} = 0$ and hence $\int_{I} (∣ a^{'} ∣^{2} - k^{2} ∣ a ∣^{2}) = \int_{I} ϕ^{2} ∣ w^{'} ∣^{2} \geq 0$ , all integrals agreeing in the Riemann and Lebesgue senses.

2.2F1F2F6F5step 1.1algebra

$ϕ \in H_{0}^{1} (I; K)$ : for integers $m \geq 4 / L$ put $η_{m} (x) : = σ (m x - 1) σ (m (L - x) - 1)$ and $ϕ_{m} : = η_{m} ϕ$ . Each factor is smooth, so $η_{m} \in C_{c}^{\infty} (I; R)$ with its support contained in $[1 / m, L - 1 / m] \subset I$ , hence $ϕ_{m} \in C_{c}^{\infty} (I; K)$ . By the chain and product rules, $η_{m}^{'} = (m σ^{'} (m x - 1)) σ (m (L - x) - 1) - σ (m x - 1) (m σ^{'} (m (L - x) - 1))$ , so $∣ η_{m}^{'} ∣ \leq 2 m C_{σ}$ , while $η_{m} = 1$ on $[2 / m, L - 2 / m]$ . Hence $ψ_{m} : = ϕ_{m} - ϕ$ is supported in $[0, 2 / m] \cup [L - 2 / m, L]$ , where $∣ ψ_{m} ∣ = ∣ (η_{m} - 1) ϕ ∣ \leq ∣ ϕ ∣ \leq 2 k / m$ by step 1.1, and $∣ ψ_{m}^{'} ∣ = ∣ η_{m}^{'} ϕ + (η_{m} - 1) ϕ^{'} ∣ \leq 2 m C_{σ} \cdot 2 k / m + k = 4 k C_{σ} + k$ there. So $∥ ψ_{m} ∥_{L^{2} (I)}^{2} \leq (2 k / m)^{2} L$ and, splitting the integral over the two strips, $∥ ψ_{m}^{'} ∥_{L^{2} (I)}^{2} \leq (4 k C_{σ} + k)^{2} \cdot 4 / m$ , both tending to $0$ . Thus $∥ ϕ_{m} - ϕ ∥_{H^{1}} \to 0$ with $ϕ_{m} \in C_{c}^{\infty} (I; K)$ , and $ϕ$ lies in the closure $H_{0}^{1} (I; K)$ .

2.3F2F3step 1.1

Weak identity on smooth tests: let $v \in C_{c}^{\infty} (I; K)$ and choose $a < b$ in $I$ with $supp v \subseteq (a, b)$ . Applying integration by parts to the real and imaginary parts of $v$ with the real function $ϕ^{'}$ gives $\int_{a}^{b} ϕ^{'} \overline{v^{'}} = [ϕ^{'} \overline{v}]_{a}^{b} - \int_{a}^{b} ϕ^{''} \overline{v} = k^{2} \int_{a}^{b} ϕ \overline{v}$ , since $ϕ^{'} v$ vanishes at $a$ and $b$ and $ϕ^{''} = - k^{2} ϕ$ by step 1.1.

3.1F1step 2.1algebra

Complex smooth case: let $u \in C_{c}^{\infty} (I; K)$ and write $u = a + i b$ with real $a, b \in C_{c}^{\infty} (I)$ (and $b = 0$ when $K = R$ ). Differentiation is componentwise, so $∣ u^{'} ∣^{2} = ∣ a^{'} ∣^{2} + ∣ b^{'} ∣^{2}$ and $∣ u ∣^{2} = a^{2} + b^{2}$ ; applying step 2.1 to $a$ and to $b$ and adding gives $\int_{I} (∣ u^{'} ∣^{2} - k^{2} ∣ u ∣^{2}) = \int_{I} ϕ^{2} (∣ w_{a}^{'} ∣^{2} + ∣ w_{b}^{'} ∣^{2}) \geq 0$ with $w_{a} = a / ϕ$ , $w_{b} = b / ϕ$ . Therefore every $u \in C_{c}^{\infty} (I; K)$ satisfies $∥ u ∥_{L^{2} (I)} \leq \frac{L}{π} ∥ u^{'} ∥_{L^{2} (I)}$ , since $k = π / L$ .

3.2F1F7step 2.3

Weak identity on $H_{0}^{1}$ : define $Λ (v) : = \int_{I} ϕ^{'} \overline{v^{'}} - k^{2} \int_{I} ϕ \overline{v}$ for $v \in H_{0}^{1} (I; K)$ . By H"older, $∣ Λ (v) ∣ \leq (∥ ϕ^{'} ∥_{2} + k^{2} ∥ ϕ ∥_{2}) ∥ v ∥_{H^{1}}$ , so $Λ$ is bounded, and it vanishes on $C_{c}^{\infty} (I; K)$ by step 2.3. For arbitrary $v \in H_{0}^{1}$ take $v_{n} \in C_{c}^{\infty} (I; K)$ with $∥ v_{n} - v ∥_{H^{1}} \to 0$ ; then $∣ Λ (v) ∣ = ∣ Λ (v) - Λ (v_{n}) ∣ \leq C ∥ v - v_{n} ∥_{H^{1}} \to 0$ , so $Λ (v) = 0$ . Hence $\int_{I} ϕ^{'} \overline{v^{'}} = k^{2} \int_{I} ϕ \overline{v}$ for every $v \in H_{0}^{1} (I; K)$ .

4.1F1F7F8step 3.1

Approximation: let $u \in H_{0}^{1} (I; K)$ . By definition of the closure there are $u_{n} \in C_{c}^{\infty} (I; K)$ with $∥ u_{n} - u ∥_{H^{1}} \to 0$ , so $∥ u_{n} - u ∥_{L^{2}} \to 0$ and $∥ u_{n}^{'} - u^{'} ∥_{L^{2}} \to 0$ . Step 3.1 gives $∥ u_{n} ∥_{2} \leq \frac{L}{π} ∥ u_{n}^{'} ∥_{2}$ for every $n$ , and the reverse triangle inequality turns both sides into convergent sequences with limits $∥ u ∥_{2}$ and $∥ u^{'} ∥_{2}$ ; passing to the limit gives $∥ u ∥_{L^{2} (I)} \leq \frac{L}{π} ∥ u^{'} ∥_{L^{2} (I)}$ .

5.1step 2.2step 1.2step 3.2algebra∎

Optimality: step 2.2 puts $ϕ$ in $H_{0}^{1} (I; K)$ and step 1.2 makes it nonzero; taking $v = ϕ$ in the identity of step 3.2 gives $∥ ϕ^{'} ∥_{L^{2}}^{2} = k^{2} ∥ ϕ ∥_{L^{2}}^{2}$ , that is $∥ ϕ ∥_{L^{2}} = \frac{L}{π} ∥ ϕ^{'} ∥_{L^{2}}$ : the constant $L / π$ is attained, hence optimal.

5 · Examples, counterexamples and false statements

None yet.

Lax--Milgram and Weak Elliptic Solutions

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Bounded, coercive and symmetric sesquilinear forms

Definition

A bounded form is represented by a unique bounded operator

Statement

Facts & Assumptions

Proof

A coercive form operator is bounded below

Statement

Facts & Assumptions

Proof

A bounded-below operator has closed range

Statement

Facts & Assumptions

Proof

The adjoint of a coercive form is coercive with the same constants

Statement

Facts & Assumptions

Proof

Coercivity of the adjoint makes the form-operator range dense

Statement

Facts & Assumptions

Proof

Coercivity makes a small form step a strict contraction

Statement

Facts & Assumptions

Proof

The Lax--Milgram theorem

Statement

Facts & Assumptions

Proof

The Lax--Milgram solution operator has norm at most 1/α1/α

Statement

Facts & Assumptions

Proof

Symmetric Lax--Milgram is energy minimisation

Statement

Facts & Assumptions

Proof

Nonsymmetric Lax--Milgram is not a scalar minimisation principle

Remark

The negative Sobolev space H−1(Ω)H−1(Ω)

Definition

L2L2 forcing and divergence data embed in H−1H−1 with a quantitative bound

Statement

Facts & Assumptions

Proof

Remarks

Every H−1H−1 functional is an L2L2 function plus a divergence

Statement

Facts & Assumptions

Proof

Uniformly elliptic divergence-form operators and their sesquilinear forms

Definition

Weak Dirichlet solutions for a divergence-form operator

Definition

The elliptic form is well defined and bounded on H1H1

Statement

Facts & Assumptions

Proof

Coercivity of the principal Dirichlet form

Statement

Facts & Assumptions

Proof

Testing a coercive weak solution with itself gives the energy bound

Statement

Facts & Assumptions

Proof

Existence and uniqueness for the weak Dirichlet Poisson problem

Statement

Facts & Assumptions

Proof

Lax--Milgram solvability for coercive divergence-form equations

Statement

Facts & Assumptions

Proof

The Lax--Milgram solution operator has norm at most $1 / α$

The negative Sobolev space $H^{- 1} (Ω)$

$L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound

Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence

The elliptic form is well defined and bounded on $H^{1}$

The Sobolev space $H^{1}$ is a Hilbert space