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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).

[L8]

The integral of a nonnegative function over a null set is zero (A nonnegative integral over a null set vanishes).

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

1.1L2L3L8given

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 [L3] makes the corresponding function vanish outside a measurable null set. By [L8], $\int ∣ f g ∣ d μ = 0$ . Thus only the case $A, B > 0$ remains.

1.2L2L4L6L7L8given

For the endpoint pair $(p, q) = (1, \infty)$ , let $M : = ∥ g ∥_{\infty}$ . By [L4] there is a measurable null set $N$ outside which $∣ f g ∣ \leq M ∣ f ∣$ . On $N$ , [L8] gives zero integral for $∣ f g ∣ χ_{N}$ and $M ∣ f ∣ χ_{N}$ . Integrating the inequality on $X ∖ N$ with [L6] and [L7] therefore gives $\int ∣ f g ∣ d μ \leq M \int ∣ f ∣ d μ = ∥ g ∥_{\infty} ∥ f ∥_{1}$ .

2.1step 1.1L1L2L5L6L7algebra

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 $\int ∣ f g ∣ d μ \leq \frac{B}{p A^{p - 1}} \int ∣ f ∣^{p} d μ + \frac{A}{q B^{q - 1}} \int ∣ g ∣^{q} d μ = \frac{A B}{p} + \frac{A B}{q} = A B .$

2.2step 1.2given

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

3.1step 2.1step 1.2step 2.2L2∎

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.

Holder's inequality for integrals, including the endpoint cases

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources