LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Real L2 kernels define bounded operators and represent rank-one tensors

Statement

Let $I = [0, 1]$ with its Borel sigma-algebra and Lebesgue measure $d t$ , and let $H = L^{2} (I, d t; R)$ and $K = L^{2} (I^{2}, d t \times d t; R)$ be the quotient spaces by null functions. Assume Countable Choice, so these $L^{2}$ spaces have their usual real Hilbert-space structures. For $k \in K$ and $f \in H$ , define, for almost every $u$ , $(T_{k} f) (u) = \int_{0}^{1} k (u, v) f (v) d v .$ This defines a bounded linear operator $T_{k} : H \to H$ and $∥ T_{k} ∥ \leq ∥ k ∥_{K}$ . For $x, y \in H$ , the class $x ⊠ y$ represented by $x (u) y (v)$ is well-defined in $K$ , satisfies $∥ x ⊠ y ∥_{K} = ∥ x ∥_{H} ∥ y ∥_{H},$ and $T_{x ⊠ y} = x \otimes y$ , where $(x \otimes y) h = ⟨ h, y ⟩_{H} x$ . For $x, y, g, h \in H$ the product-kernel inner product also satisfies $⟨ x ⊠ y, g ⊠ h ⟩_{K} = ⟨ x, g ⟩_{H} ⟨ y, h ⟩_{H} .$

Facts & Assumptions

Given: Countable Choice; the Borel probability space $(I, B, d t)$ ; the product measure $d t \times d t$ on $I^{2}$ ; and the real $L^{2}$ quotient spaces $H$ and $K$ .

[A1]

Countable Choice is assumed for the Hilbert-space $L^{2}$ interfaces (The Axiom of Countable Choice ( ${A C}_{ω}$ ), $L^{2}$ with the integral pairing is a Hilbert space).

[F1]

The product measure is sigma-finite and satisfies $(d t \times d t) (A \times B) = d t (A) d t (B)$ for Borel $A, B$ (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, Finite, sigma-finite, and semifinite measures).

[F2]

Tonelli applies to nonnegative Borel functions on $I^{2}$ and makes their section integrals measurable (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F3]

If a product-measurable function is in $L^{1} (d t \times d t)$ , its sections are integrable almost everywhere, the section-integral function is measurable after zero extension on the exceptional null set, and its integral equals the product integral (Fubini's theorem for L^1 functions on a sigma-finite product).

[F4]

The real $L^{2}$ quotient has inner product $⟨ f, g ⟩ = \int f g d t$ and satisfies Cauchy–Schwarz ( $L^{2}$ with the integral pairing is a Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions, Real and complex inner-product spaces and their induced length).

Proof

Given: The spaces, functions, and conventions in the Statement and Facts & Assumptions.

1.1A1F1F2F4

Choose measurable representatives of $x$ and $y$ . By [F1]–[F2], $\int_{I^{2}} ∣ x (u) y (v) ∣^{2} d (d t \times d t) = (\int_{I} ∣ x (u) ∣^{2} d u) (\int_{I} ∣ y (v) ∣^{2} d v) < \infty .$ Changing either representative changes the product only on $(N \times I) \cup (I \times M)$ for null sets $N, M \subset I$ ; [F1] gives this union product measure zero. Hence $x ⊠ y$ is a well-defined class in $K$ and the displayed integral proves its norm identity.

1.2F1F2F3F4algebra

Choose Borel representatives of $k$ and $f$ . The function $(u, v) \mapsto k (u, v) f (v)$ is product-measurable and belongs to $L^{1} (d t \times d t)$ , since Cauchy–Schwarz on $I^{2}$ gives $\int_{I^{2}} ∣ k (u, v) f (v) ∣ d u d v \leq ∥ k ∥_{K} {(\int_{I^{2}} ∣ f (v) ∣^{2} d u d v)}^{1 / 2} = ∥ k ∥_{K} ∥ f ∥_{H} .$ By [F3] its section integrals are defined almost everywhere and have a measurable representative. For almost every $u$ , sectionwise Cauchy–Schwarz and [F2] give $∣ (T_{k} f) (u) ∣^{2} \leq ∥ f ∥_{H}^{2} \int_{I} ∣ k (u, v) ∣^{2} d v, ∥ T_{k} f ∥_{H}^{2} \leq ∥ k ∥_{K}^{2} ∥ f ∥_{H}^{2} .$ If $f$ is changed on a $d t$ -null set, its section integral is unchanged for every $u$ whose $k (u, \cdot)$ is square-integrable. If $k$ is changed on a $(d t \times d t)$ -null set, the difference times $f (v)$ is in $L^{1}$ by the same Cauchy–Schwarz estimate, so [F3] makes its section integral zero for almost every $u$ . Thus $T_{k}$ is well-defined on the quotients. Linearity follows from linearity of the section integral, and the second estimate proves boundedness and $∥ T_{k} ∥ \leq ∥ k ∥_{K}$ .

2.1step 1.1step 1.2F4algebra

For $k = x ⊠ y$ and almost every $u$ , the section is integrable by step 1.2 and $(T_{x ⊠ y} f) (u) = x (u) \int_{I} y (v) f (v) d v = x (u) ⟨ f, y ⟩_{H} = ((x \otimes y) f) (u) .$ Thus $T_{x ⊠ y} = x \otimes y$ as operators on $H$ . This includes $x = 0$ or $y = 0$ , when both operators are zero.

3.1F1F3F4step 1.1algebra∎

For $x, y, g, h \in H$ , [F3] applies to $x (u) y (v) g (u) h (v)$ because Cauchy–Schwarz on $I^{2}$ bounds its absolute integral by $∥ x ∥_{H} ∥ g ∥_{H} ∥ y ∥_{H} ∥ h ∥_{H}$ . Fubini then factors the integral into $⟨ x, g ⟩_{H} ⟨ y, h ⟩_{H}$ , proving the stated inner-product identity.

Depends on

Used by

Long-run covariance of an L2-m-approximable shift is trace class Lemma

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.