PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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The $p$ -functional need not be a norm for $0 < p < 1$

Statement

Let $0 < p < 1$ . Then the functional

$[f] ⟼ {(\int ∣ f ∣^{p} d μ)}^{1 / p}$

on $L^{p} (μ)$ need not satisfy the triangle inequality. Consequently it is not a norm in general in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms.

Facts & Assumptions

Given: A real exponent $0 < p < 1$ .

[L1]

Real powers are defined for positive bases and obey the exponent laws; for fixed base $2 > 1$ , the map $x \mapsto 2^{x}$ is strictly increasing (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential function is strictly increasing).

[L2]

A norm must satisfy the triangle inequality (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

Counting measure on a two-point set is a measure (Counting measure on an arbitrary set, Counting measure is a measure).

Proof

Proof technique: Use two disjoint equal-mass indicators, so the triangle inequality becomes the scalar inequality $2^{1 / p} \leq 2$ , which fails because $1 / p > 1$ .

1.1

Work on the two-point counting space ${0, 1}$ from [L3]. Let [L3, given] $e_{0} : = χ_{{0}}$ and $e_{1} : = χ_{{1}}$ . Then $∥ e_{0} ∥_{p} = ∥ e_{1} ∥_{p} = 1, ∥ e_{0} + e_{1} ∥_{p} = (1 + 1)^{1 / p} = 2^{1 / p} .$

2.1

Because $0 < p < 1$ , one has $1 / p > 1$ . Strict monotonicity in [L1] therefore [L1, step 1.1] gives $2^{1 / p} > 2^{1} = 2.$ So $∥ e_{0} + e_{1} ∥_{p} > ∥ e_{0} ∥_{p} + ∥ e_{1} ∥_{p} .$

3.1

The triangle inequality from [L2] fails on this concrete measure space, so [L2, step 2.1] the $p$ -functional is not a norm in general for $0 < p < 1$ . ∎

Depends on

Used by

The half-norm fails the triangle inequality on two indicators Counterexample

Dependency tree · two levels

37 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, Theorem 8.16 (standard reference, not scraped)
John K. Hunter, Measure Theory, reverse inequality discussion before Definition 7.6 (standard reference, not scraped)

The pp-functional need not be a norm for 0<p<10<p<1