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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 $0 < p < \infty$ ).

[L5]

For a weight strictly between zero and one, equality in the two-point exponential convexity inequality holds exactly when its two inputs agree (The two-point convexity inequality for the exponential function).

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

1.1L1L3L4

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 $0$ .

1.2L1L2L3

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 $v = ∣ g ∣ / B$ .

2.1L2L5step 1.2algebra

To check the equality condition omitted from [L2]'s Statement, first suppose $u, v > 0$ . Apply [L5] to $\log (u^{p})$ and $\log (v^{q})$ with weights $1 / p$ and $1 / q$ : its geometric side is $u v$ , its arithmetic side is $u^{p} / p + v^{q} / q$ , and equality holds exactly when $u^{p} = v^{q}$ . If one of $u, v$ is zero, equality in Young holds only when both vanish, again exactly when $u^{p} = v^{q}$ . Applying this criterion to step 1.2 gives [step 1.2, L2, L5, algebra] $\frac{∣ f ∣^{p}}{A^{p}} = \frac{∣ g ∣^{q}}{B^{q}} μ -almost everywhere,$ so $∣ f ∣^{p} = (A^{p} / B^{q}) ∣ g ∣^{q}$ almost everywhere.

3.1step 1.1step 2.1L1L4algebra

Conversely, if $∣ f ∣^{p} = c ∣ g ∣^{q}$ almost everywhere for some $c > 0$ and neither function vanishes almost everywhere, integration gives $A^{p} = c B^{q}$ . Thus the normalized powers agree almost everywhere, so the scalar criterion from step 2.1 makes Young an equality almost everywhere and the integrated Holder proof becomes an equality. If either function vanishes almost everywhere, step 1.1 already applies.

4.1step 1.1step 2.1step 3.1∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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