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Holder's inequality for integrals, including the endpoint cases

Statement

Let $(X, A, μ)$ be a measure space, let $p, q \in [1, \infty]$ be conjugate exponents, and let $f, g$ be measurable real-valued functions.

If $1 \leq p < \infty$ and $q < \infty$ with $f \in L^{p} (μ)$ and $g \in L^{q} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{p} ∥ g ∥_{q} .$
If $p = 1$ and $q = \infty$ with $f \in L^{1} (μ)$ and $g \in L^{\infty} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{1} ∥ g ∥_{\infty} .$
If $p = \infty$ and $q = 1$ with $f \in L^{\infty} (μ)$ and $g \in L^{1} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{\infty} ∥ g ∥_{1} .$

In every case the right-hand side is finite, so $f g$ is integrable.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , conjugate exponents $p, q \in [1, \infty]$ , and measurable real-valued functions $f, g$ in the spaces named in the relevant clause of the Statement.

[L1]

Conjugate exponents are defined in Conjugate exponents, including the endpoint conventions.

[L2]

For $0 < r < \infty$ , membership in $L^{r} (μ)$ means $\int ∣ h ∣^{r} d μ < \infty$ , while $L^{\infty} (μ)$ means finite essential supremum (The function space $L^{p} (μ)$ for $0 < p < \infty$ , The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[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]

If $h \in L^{\infty} (μ)$ , then $∣ h ∣ \leq ∥ h ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

[L5]

Young's inequality says $u v \leq u^{p} / p + v^{q} / q$ for $u, v \geq 0$ when $1 < p, q < \infty$ are conjugate (Young's inequality for conjugate real exponents).

[L6]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L7]

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

Proof

technique · For $1 < p < infinity$, normalize the nonzero norms and apply the published Young inequality pointwise before integrating. Treat the endpoint pairs $(1,\infty)$ and $(\infty,1)$ separately from the essential-bound definition

1.1

Assume first $1 < p, q < \infty$ , and put $A : = ∥ f ∥_{p}$ and $B : = ∥ g ∥_{q}$ . If $A = 0$ or $B = 0$ , then the corresponding power integral is $0$ , so the corresponding function vanishes almost everywhere and $\int ∣ f g ∣ d μ = 0$ . Thus only the case $A, B > 0$ remains.

L2L3given

1.2

For the endpoint pair $(p, q) = (1, \infty)$ , let $M : = ∥ g ∥_{\infty}$ . Then [L2, L4, L6, given] $\int ∣ f g ∣ d μ \leq M \int ∣ f ∣ d μ = ∥ g ∥_{\infty} ∥ f ∥_{1} .$ Indeed, [L4] gives a measurable null set $N$ with $∣ g ∣ \leq M$ on $X ∖ N$ , so $∣ f g ∣ \leq M ∣ f ∣$ almost everywhere.

2.1

In the remaining strict-exponent case, Young's inequality applied pointwise to $u = ∣ f ∣ / A$ and $v = ∣ g ∣ / B$ gives [step 1.1, L1, L2, L5, L6, L7, algebra] $\frac{∣ f ∣ ∣ g ∣}{A B} \leq \frac{∣ f ∣^{p}}{p A^{p}} + \frac{∣ g ∣^{q}}{q B^{q}} .$ Integrating and using additivity, monotonicity, homogeneity, and the definitions of $A$ and $B$ yields $\int ∣ f g ∣ d μ \leq \frac{B}{p A^{p - 1}} \int ∣ f ∣^{p} d μ + \frac{A}{q B^{q - 1}} \int ∣ g ∣^{q} d μ = \frac{A B}{p} + \frac{A B}{q} = A B .$

2.2

The case $(p, q) = (\infty, 1)$ is identical after exchanging $f$ and $g$ . [step 1.2, given] $\int ∣ f g ∣ d μ \leq ∥ f ∥_{\infty} ∥ g ∥_{1} .$

3.1

Step 2.1 proves the strict-exponent case, and steps 1.2 and 2.2 prove the two endpoint cases. In every case the right-hand side is finite by [L2], so $f g$ is integrable.

step 2.1step 1.2step 2.2L2∎

Depends on

Used by

Cauchy-Schwarz inequality for L² Corollary
FALSE: Holder equality forces the functions themselves to be proportional False statement
Finite counting measure recovers finite Holder and implies the signed Cauchy-Schwarz inequality Remark
Equality in Holder's inequality for 1 < p < ∞ Theorem
Finite-measure Lʳ includes into Lᵖ for p < r Theorem
Generalized Holder inequality puts products into Lʳ Theorem
Lyapunov interpolation inequality for Lᵖ norms Theorem
Minkowski's inequality for integrals, including p = ∞ Theorem

Dependency tree · two levels

25 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)
John K. Hunter, Measure Theory, Section 7.2 (standard reference, not scraped)