False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

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FALSE: the $p$ -seminorm on calligraphic $L^{p}$ is a norm

Statement

On the representative space $L^{p} (μ)$ , the functional $f \mapsto ∥ f ∥_{p}$ is a norm.

Facts & Assumptions

Given: A nonzero function with zero seminorm.

[L1]

The previous counterexample supplies a measurable function $f \equiv̸ 0$ with $∥ f ∥_{p} = 0$ (A nonzero function on a null set has zero $L^{p}$ seminorm).

[L2]

A norm must satisfy the separation axiom $N (v) = 0 \Leftrightarrow v = 0$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Refutation

Proof technique: Refute with a nonzero function supported on a null set, whose seminorm is $0$ .

1.1

Let $f$ be the function from [L1]. Then $f \neq 0$ pointwise but [L1] $∥ f ∥_{p} = 0$ .

2.1

This violates the separation axiom in [L2], so the $p$ -seminorm on [step 1.1, L2] $L^{p} (μ)$ is not a norm. ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Sheldon Axler, Measure, Integration & Real Analysis, Section 7B (standard reference, not scraped)

FALSE: the pp-seminorm on calligraphic LpLp is a norm