CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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.

The inhomogeneous weak Dirichlet problem by a trace lifting

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources