CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-09-01

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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.

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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 $f (x) = g (x) : = (1 + ∣ x ∣)^{- 3 / 4} .$

[L1]

Young's inequality controls only the exponents satisfying $1 / r = 1 / p + 1 / q - 1$ (Young's convolution inequality).

Counterexample

technique · direct

1.1

Since [L1, given, algebra] $\int_{R} (1 + ∣ x ∣)^{- 3 / 2} d x < \infty,$ the functions $f$ and $g$ lie in $L^{2} (R)$ .

L1givenalgebra

2.1

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 $c > 0$ .

step 1.1algebra

3.1

But [L1, step 2.1] $\int_{2}^{\infty} x^{- 1} d x = \infty,$ so the lower bound from step 2.1 shows $f * g \notin L^{2} (R)$ . Hence the statement refuted above is false.

L1step 2.1∎

Depends on

Young's convolution inequality

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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.

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