LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Locally integrable functions define distributions

Statement

Let $Ω \subseteq R^{n}$ be open and let $f \in L_{l o c}^{1} (Ω; C)$ . The regular functional $T_{f}$ defined by $⟨ T_{f}, φ ⟩ = \int_{Ω} f (x) φ (x) d x (φ \in D (Ω))$ is a distribution on $Ω$ with the LF test-function topology.

Statement references

regular functional from a locally integrable function

LF test-function topology universal property

Facts & Assumptions

Given: the open set $Ω$ , the locally integrable function $f$ , and the regular functional from the Statement.

[F1]

Regular distribution from a locally integrable function defines $⟨ T_{f}, φ ⟩ = \int_{Ω} f (x) φ (x) d x$ . If $K = supp φ$ , it gives $∣ ⟨ T_{f}, φ ⟩ ∣ \leq \sup_{K} ∣ φ ∣ \int_{K} ∣ f ∣$ and states that the functional is complex-linear and well-defined.

[F2]

Test function lf topology universal property states that a linear map from $D (Ω)$ to a locally convex space is continuous if and only if its restriction to every fixed-support space $D_{K}$ is continuous. This holds in ZF.

Proof

1.1givenF1algebra

The functional $T_{f}$ is linear by [F1]. For each compact $K \subseteq Ω$ , [F1] gives $∣ ⟨ T_{f}, φ ⟩ ∣ \leq (\int_{K} ∣ f ∣) p_{0, K} (φ)$ for every $φ \in D_{K}$ , where $p_{0, K} (φ) = \sup_{x \in K} ∣ φ (x) ∣$ and the value is $0$ when $K = \emptyset$ . Since $f$ is locally integrable, the coefficient is finite, so the restriction $T_{f} ∣_{D_{K}}$ is continuous.

2.1givenF1F2step 1.1∎

Apply [F2] to the linear map $T_{f}$ : every fixed-support restriction is continuous by step 1.1, so $T_{f}$ is continuous on $D (Ω)$ with its LF topology. A continuous complex-linear functional on this space is a distribution, as claimed. The proof uses no choice axiom and asserts no injectivity result.

Depends on

Dependency tree · 2 levels

Sources

Razvan Gelca, Functional Analysis, lecture notes, Chapter 7, Section 7.3