CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

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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.

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$L^{2}$ is not a subset of $L^{1}$ on the line

Statement refuted

On $R$ with Lebesgue measure, every $L^{2}$ function lies in $L^{1}$ .

Facts & Assumptions

Given: The power-family thresholds from $x^{- a}$ on $(0, 1)$ and on $(1, \infty)$ calibrates $L^{p}$ membership.

[L1]

For $f_{\infty} (x) = x^{- a} χ_{(1, \infty)} (x)$ , one has $f_{\infty} \in L^{p}$ exactly when $a p > 1$ ( $x^{- a}$ on $(0, 1)$ and on $(1, \infty)$ calibrates $L^{p}$ membership).

Counterexample

Proof technique: Use the power-function family $x^{- a}$ at infinity with $1 / 2 < a \leq 1$ .

1.1

Choose $a = 3 / 4$ and set $f (x) : = x^{- 3 / 4} χ_{(1, \infty)} (x)$ . Then [L1] $a \cdot 2 = 3 / 2 > 1$ , so [L1] gives $f \in L^{2}$ . But $a \cdot 1 = 3 / 4 ≯ 1$ , so gives $f \notin L^{1}$ .

L1given

2.1

Thus $f$ is an $L^{2}$ function on $R$ that does not lie in $L^{1}$ , [step 1.1] refuting the claim. ∎

Depends on

$x^{-a}$ on $(0,1)$ and on $(1,\infty)$ calibrates $L^p$ membership

Used by

FALSE: Lᵖ includes into Lʳ on every measure space when p < r False statement

Dependency tree · two levels

5 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, Chapter 8 (standard reference, not scraped)

L2L2 is not a subset of L1L1 on the line