TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Free Laplacian form domain and realization

Statement

Assume the Axiom of Choice and Countable Choice. On $H = L^{2} (R^{3}; C)$ use the unitary Fourier transform $\hat{u} (ξ) = \int_{R^{3}} e^{- 2 π i x \cdot ξ} u (x) d x$ on $L^{1} \cap L^{2}$ . Let $H_{0} = - Δ$ be the self-adjoint multiplier $F_{2}^{*} M_{4 π^{2} ∣ ξ ∣^{2}} F_{2}$ with domain $D (H_{0}) = W^{2, 2} (R^{3}; C)$ . Put $Q_{0} = W^{1, 2} (R^{3}; C)$ and $q_{0} (u, v) = \sum_{j = 1}^{3} ⟨ \partial_{j} u, \partial_{j} v ⟩_{L^{2}} (u, v \in Q_{0}),$ where $\partial_{j}$ are weak derivatives. Then $q_{0}$ is dense, symmetric, nonnegative, and closed; its associated self-adjoint operator is exactly $H_{0}$ , with $Q (H_{0}) = D (H_{0}^{1 / 2}) = Q_{0}$ and $q_{0} (u, v) = ⟨ H_{0} u, v ⟩ (u \in D (H_{0}), v \in Q_{0}) .$

Statement references

Real multiplication operators and free plus bounded-potential Schrödinger operators

Integer-order Sobolev spaces and their norms

Semibounded closed-form representation and perturbation

Facts & Assumptions

Given: Choice and Countable Choice, $H = L^{2} (R^{3}; C)$ , $F_{2}$ with the stated $2 π$ convention, and weak-derivative spaces $W^{1, 2}$ and $W^{2, 2}$ .

[A1]

Choice and Countable Choice are the hypotheses of the cited free multiplier, Fourier-Sobolev and Plancherel suppliers (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The free operator is self-adjoint with $H_{0} = F_{2}^{*} M_{4 π^{2} ∣ ξ ∣^{2}} F_{2}$ and $D (H_{0}) = W^{2, 2} (R^{3}; C)$ ; it is nonnegative (Real multiplication operators and free plus bounded-potential Schrödinger operators).

[F2]

$F_{2}$ is unitary, uses $\hat{u} (ξ) = \int e^{- 2 π i x \cdot ξ} u (x) d x$ on $L^{1} \cap L^{2}$ , and $F_{2} (\partial_{j} u) = 2 π i ξ_{j} F_{2} u$ for weak derivatives (Plancherel theorem, Integer-order W^{k,2} and H^k agree with equivalent norms).

[F3]

$W^{k, 2}$ consists of $L^{2}$ classes whose weak derivatives through order $k$ lie in $L^{2}$ ; the Fourier characterization identifies $W^{k, 2}$ with the corresponding weighted $L^{2}$ multiplier domain and makes its derivative norm complete (Integer-order Sobolev spaces and their norms, Integer-order W^{k,2} and H^k agree with equivalent norms).

[F4]

A symmetric form is closed when its shifted form norm is complete; a self-adjoint operator represented by such a form is unique and has that form domain equal to its square-root domain (Closed semibounded symmetric forms and their associated operators, Semibounded closed-form representation and perturbation).

[F5]

Schwartz functions are dense in $L^{2} (R^{3}; C)$ and belong to every integer-order Sobolev space, and Fourier transformation maps Schwartz space onto itself (Schwartz space is dense in L2, Integer-order W^{k,2} and H^k agree with equivalent norms, Fourier transform is a topological automorphism of Schwartz space).

[F6]

The $L^{2}$ inner product satisfies Cauchy–Schwarz (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F7]

The regular distribution map on locally integrable functions is injective modulo almost-everywhere equality; its test functions include $C_{c}^{\infty} (R^{3})$ , which is contained in Schwartz space (Locally integrable functions embed in distributions).

Proof

Given: [A1] and the Hilbert space, Fourier transform, weak derivatives, and free multiplier stated above.

1.1A1F2F3F4F5algebra

Define $q_{0}$ on $Q_{0} = W^{1, 2}$ by the displayed weak-derivative pairing. By [F2], for $u, v \in Q_{0}$ , $q_{0} (u, v) = \int_{R^{3}} 4 π^{2} ∣ ξ ∣^{2} \hat{u} (ξ) \overline{\hat{v} (ξ)} d ξ .$ Thus $q_{0}$ is sesquilinear, symmetric and nonnegative. Moreover $q_{0} [u] + ∥ u ∥_{2}^{2} = ∥ u ∥_{W^{1, 2}}^{2}$ , which is complete by [F3], so $q_{0}$ is closed by [F4]. Since Schwartz functions lie in $Q_{0}$ and are dense in $L^{2}$ by [F5], $Q_{0}$ is dense.

1.2F1F2F3F6given

If $u \in D (H_{0}) = W^{2, 2}$ and $v \in Q_{0}$ , Plancherel [F2] and Cauchy–Schwarz [F6] give $q_{0} (u, v) = \int 4 π^{2} ∣ ξ ∣^{2} \hat{u} \overline{\hat{v}} = ⟨ H_{0} u, v ⟩$ ; both factors in this pairing are in $L^{2}$ by the $W^{2, 2}$ and $W^{1, 2}$ hypotheses.

1.3F1F2F3F5F7given

Conversely, suppose $u \in Q_{0}$ and $f \in H$ satisfy $q_{0} (u, v) = ⟨ f, v ⟩$ for every $v \in Q_{0}$ . Testing with Schwartz $v$ and using Plancherel gives $\int (4 π^{2} ∣ ξ ∣^{2} \hat{u} (ξ) - \hat{f} (ξ)) \overline{\hat{v} (ξ)} d ξ = 0$ . The Fourier transform maps Schwartz space onto itself by [F5], and complex conjugation preserves Schwartz space, so the locally integrable function $4 π^{2} ∣ ξ ∣^{2} \hat{u} - \hat{f}$ pairs to zero with every compactly supported smooth test function. It is locally integrable because $\hat{u}, \hat{f} \in L^{2}$ and the polynomial weight is bounded on compact sets. Injectivity of the regular-distribution map [F7] gives $4 π^{2} ∣ ξ ∣^{2} \hat{u} = \hat{f}$ almost everywhere; hence $∣ ξ ∣^{2} \hat{u} \in L^{2}$ . By [F3], $u \in W^{2, 2} = D (H_{0})$ , and [F1] then gives $H_{0} u = f$ . Thus the form-associated operator domain and action are exactly those of $H_{0}$ .

2.1F1F4step 1.1step 1.2step 1.3∎

Steps 1.2-1.3 show that the nonnegative self-adjoint operator $H_{0}$ is the operator represented by the dense closed form $q_{0}$ . Uniqueness in [F4] therefore gives $Q (H_{0}) = D (H_{0}^{1 / 2}) = Q_{0}$ , completing the Statement.

Free Laplacian form domain and realization

Statement

Statement references

Facts & Assumptions

Proof

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