False statementConstruction: 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.
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FALSE: if $f, g \in L^{1} (R^{n})$ , then $f * g (x)$ is defined for every $x$

Statement

False claim. If $f, g \in L^{1} (R^{n})$ , then $(f * g) (x)$ is defined for every $x \in R^{n}$ .

Facts & Assumptions

Given: The one-dimensional functions $f (x) = g (x) : = {\begin{cases} \frac{1}{∣ x ∣ (\log (e / ∣ x ∣))^{2}}, & 0 < ∣ x ∣ < 1 / 2, \\ 0, & otherwise . \end{cases}$

[L1]

The $L^{1}$ convolution theorem guarantees only almost-everywhere existence (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ ).

Refutation

technique · direct

1.1

The function $x \mapsto 1 / (∣ x ∣ (\log (e / ∣ x ∣))^{2})$ is integrable near $0$ , so [L1, given, algebra] $f, g \in L^{1} (R)$ .

L1givenalgebra

2.1

At $x = 0$ one has [step 1.1, algebra] $\int_{R} ∣ f (- y) g (y) ∣ d y = \int_{- 1 / 2}^{1 / 2} \frac{d y}{∣ y ∣^{2} (\log (e / ∣ y ∣))^{4}} .$ The single point $y = 0$ is irrelevant to Lebesgue integrability, so it is enough to inspect the punctured interval $(0, 1 / 2)$ . There the substitution $u = \log (e / y)$ gives $y = e^{1 - u}$ and $d y = - e^{1 - u} d u$ , hence $\int_{0}^{1 / 2} \frac{d y}{y^{2} (\log (e / y))^{4}} = \int_{\log (2 e)}^{\infty} \frac{e^{u - 1}}{u^{4}} d u = \infty .$ Thus $(f * g) (0)$ is not defined as an absolutely convergent Lebesgue integral.

step 1.1algebra

3.1

Therefore the convolution of two $L^{1}$ functions need not be defined at [L1, step 2.1] every point; [L1] correctly states only almost-everywhere existence.

L1step 2.1∎

Depends on

If $f,g \in L^1(\mathbb{R}^n)$, then $f*g$ exists almost everywhere, belongs to $L^1$, and $\|f*g\|_1 \le \|f\|_1 \|g\|_1$

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

Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)

FALSE: if f,g∈L1(Rn)f,g∈L1(Rn), then f∗g(x)f∗g(x) is defined for every xx