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

Statement

Let $1 < p < \infty$ , let $q$ be its conjugate exponent, and let $f \in L^{p} (μ)$ and $g \in L^{q} (μ)$ . Then equality holds in Holder's inequality

$\int ∣ f g ∣ d μ = ∥ f ∥_{p} ∥ g ∥_{q}$

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

$∣ f ∣^{p} = c ∣ g ∣^{q} μ -almost everywhere .$

Facts & Assumptions

Given: A measure space, an exponent $1 < p < \infty$ , its conjugate exponent $q$ , and functions $f \in L^{p} (μ)$ , $g \in L^{q} (μ)$ .

[L1]

Holder's inequality for integrals has already been proved (Holder's inequality for integrals, including the endpoint cases).

[L2]

Young's inequality is the scalar step used in that proof (Young's inequality for conjugate real exponents).

[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]

Membership in $L^{p} (μ)$ and $L^{q} (μ)$ means finiteness of the corresponding power integrals (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

Proof

Proof technique: Trace where equality can occur in the normalized Young-inequality proof. Equality in Young forces the normalized powers $∣ f ∣^{p}$ and $∣ g ∣^{q}$ to be proportional almost everywhere, and conversely that proportionality makes the inequality an equality.

1.1

If $∥ f ∥_{p} = 0$ or $∥ g ∥_{q} = 0$ , then the corresponding function is zero almost everywhere, and Holder's inequality becomes equality with both sides $0$ .

L1L3L4

1.2

Assume now that $A : = ∥ f ∥_{p} > 0$ and $B : = ∥ g ∥_{q} > 0$ . The proof of [L1] integrated the nonnegative function [L1, L2, L3] $H : = \frac{∣ f ∣^{p}}{p A^{p}} + \frac{∣ g ∣^{q}}{q B^{q}} - \frac{∣ f g ∣}{A B} .$ If equality holds in Holder, then $\int H d μ = 0$ , so $H = 0$ almost everywhere. Thus equality holds in Young's inequality pointwise almost everywhere for $u = ∣ f ∣ / A$ and $v = ∣ g ∣ / B$ .

2.1

Equality in Young's inequality for conjugate exponents means $u^{p} = v^{q}$ . Applying that to step 1.2 gives [step 1.2, L2] $\frac{∣ f ∣^{p}}{A^{p}} = \frac{∣ g ∣^{q}}{B^{q}} μ -almost everywhere,$ so $∣ f ∣^{p} = (A^{p} / B^{q}) ∣ g ∣^{q}$ almost everywhere.

3.1

Conversely, if $∣ f ∣^{p} = c ∣ g ∣^{q}$ almost everywhere for some $c > 0$ , then after normalizing by the two norms the two sides in Young's inequality agree almost everywhere, so the integrated Holder proof becomes an equality.

step 2.1L1L2L4algebra

4.1

Step 1.1 handles the zero-function case, step 2.1 proves the strict necessity, and step 3.1 proves sufficiency. These are exactly the alternatives in the Statement.

step 1.1step 2.1step 3.1∎

Depends on

Used by

Cauchy-Schwarz inequality for L² Corollary
Proportional functions realise the equality case of Holder Example
FALSE: Holder equality forces the functions themselves to be proportional False statement
Equality in Minkowski's inequality for 1 < p < ∞ Theorem

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, Holder's Inequality (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem 8.6 and converse (standard reference, not scraped)

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