CounterexampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)

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.

Two $L^{2}$ functions can have convolution outside $L^{2}$

Statement refuted

Every convolution of two $L^{2} (R)$ functions again belongs to $L^{2} (R)$ .

Facts & Assumptions

Given: The one-dimensional function $f (x) = g (x) : = (1 + ∣ x ∣)^{- 3 / 4} .$

Counterexample

technique · direct

1.1givenalgebra

Since [given, algebra] $\int_{R} (1 + ∣ x ∣)^{- 3 / 2} d x < \infty,$ the functions $f$ and $g$ lie in $L^{2} (R)$ .

2.1step 1.1algebra

For each fixed $x$ , the convolution integral is finite: on a bounded $y$ -interval its continuous integrand is bounded, while for $∣ y ∣ > 2 ∣ x ∣ + 2$ one has $1 + ∣ x - y ∣ \geq (1 + ∣ y ∣) / 2$ , so the integrand is at most $2^{3 / 4} (1 + ∣ y ∣)^{- 3 / 2}$ , an integrable tail. For $x > 2$ , [step 1.1, algebra] $(f * g) (x) = \int_{R} (1 + ∣ x - y ∣)^{- 3 / 4} (1 + ∣ y ∣)^{- 3 / 4} d y \geq \int_{1}^{x - 1} x^{- 3 / 4} x^{- 3 / 4} d y = \frac{x - 2}{x^{3 / 2}} .$ So for large $x$ , $(f * g) (x) \geq c x^{- 1 / 2}$ for some $c > 0$ .

3.1step 2.1algebra∎

But [step 2.1, algebra] $\int_{2}^{\infty} x^{- 1} d x = \infty,$ so the lower bound from step 2.1 shows $f * g \notin L^{2} (R)$ . Hence the statement refuted above is false.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

Two L2L2 functions can have convolution outside L2L2