TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-31

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Equality in Minkowski's inequality for $1 < p < \infty$

Statement

Let $1 < p < \infty$ and let $f, g \in L^{p} (μ)$ . Then equality holds in Minkowski's inequality

$∥ f + g ∥_{p} = ∥ f ∥_{p} + ∥ g ∥_{p}$

if and only if at least one of $f, g$ is zero almost everywhere, or there is a constant $λ > 0$ such that

$f = λ g μ -almost everywhere .$

Facts & Assumptions

Given: An exponent $1 < p < \infty$ and functions $f, g \in L^{p} (μ)$ .

[L1]

Minkowski's inequality has already been proved (Minkowski's inequality for integrals, including $p = \infty$ ).

[L2]

The equality case in Holder has already been proved (Equality in Holder's inequality for $1 < p < \infty$ ).

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L4]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

Proof

technique · Examine the Holder step in the standard proof of Minkowski. Equality forces the nonnegative functions $|f|$ and $|g|$ to be proportional almost everywhere, and the pointwise triangle inequality then forces the same sign

1.1L1given

If at least one of $f, g$ is zero almost everywhere, then equality is immediate.

1.2L1given

If $f = λ g$ almost everywhere for some $λ > 0$ , then [L1, given] $∣ f + g ∣ = (λ + 1) ∣ g ∣$ almost everywhere, so $∥ f + g ∥_{p} = (λ + 1) ∥ g ∥_{p} = ∥ f ∥_{p} + ∥ g ∥_{p} .$

1.3L1L2L3L4

Conversely, assume equality in Minkowski and that neither $f$ nor $g$ is zero almost everywhere. [L1, L2, L3, L4] The proof of [L1] showed that equality in Minkowski can only occur when both inequalities $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣$ and $\int ∣ f ∣ ∣ f + g ∣^{p - 1} d μ \leq ∥ f ∥_{p} ∥ f + g ∥_{p}^{p - 1}$ and its $g$ -analogue are equalities. The second and third equalities force the pairs $(∣ f ∣, ∣ f + g ∣^{p - 1})$ and $(∣ g ∣, ∣ f + g ∣^{p - 1})$ to satisfy Holder equality. By [L2], this makes $∣ f ∣$ and $∣ g ∣$ proportional almost everywhere. The first inequality then forces $(∣ f ∣ + ∣ g ∣) - ∣ f + g ∣$ to have integral $0$ ; [L3] and [L4] therefore give $∣ f + g ∣ = ∣ f ∣ + ∣ g ∣ μ -almost everywhere .$

2.1step 1.3algebra

Let $∣ f ∣ = c ∣ g ∣$ almost everywhere with $c > 0$ . Then [step 1.3, algebra] on the set where $g \neq 0$ , step 1.3 gives equality in the real triangle inequality for $f$ and $g$ , so they have the same sign there. Hence $f = c g$ almost everywhere on ${g \neq 0}$ , and on ${g = 0}$ both sides vanish. Thus $f = λ g$ almost everywhere for $λ = c > 0$ .

3.1step 1.1step 1.2step 1.3step 2.1∎

Steps 1.1 and 1.2 prove sufficiency, while steps 1.3 and 2.1 prove necessity.

Depends on

Used by

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Dependency tree · two levels

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Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Chapter 8 (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 7.2 (standard reference, not scraped)

Equality in Minkowski's inequality for 1<p<∞1<p<∞