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Minkowski's integral inequality

Statement

Let $(X, μ)$ and $(Y, ν)$ be sigma-finite measure spaces, let $1 \leq p < \infty$ , and let $F : X \times Y \to C$ be measurable with

$\int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) < \infty .$

Then the function $H (x) : = \int_{Y} ∣ F (x, y) ∣ d ν (y)$ belongs to $L^{p} (X)$ and

$∥ H ∥_{L^{p} (X)} \leq \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$

Facts & Assumptions

Given: Sigma-finite measure spaces, an exponent $1 \leq p < \infty$ , and a measurable function $F$ satisfying the displayed integrability hypothesis.

[L1]

For $1 < p < \infty$ , the nonnegative duality formula is available (For $1 < p < \infty$ , the $L^{p}$ norm of a nonnegative function is the supremum of its pairings with $L^{q}$ unit vectors).

[L2]

Tonelli applies to nonnegative measurable functions on product spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[L3]

Holder's inequality is available (Holder's inequality for integrals, including the endpoint cases).

[L4]

Monotone convergence is available for nonnegative measurable functions (Monotone convergence for the integral).

Proof

technique · direct

1.1L2givenalgebra

If $p = 1$ , then Tonelli directly gives [L2, given, algebra] $∥ H ∥_{L^{1} (X)} = \int_{X} \int_{Y} ∣ F (x, y) ∣ d ν (y) d μ (x) = \int_{Y} ∥ F (\cdot, y) ∥_{L^{1} (X)} d ν (y) .$

1.2L2L3givenalgebra

Assume $1 < p < \infty$ and let $q$ be conjugate to $p$ . Put [L2, L3, given, algebra] $M : = \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$ For every nonnegative $g \in L^{q} (X)$ with $∥ g ∥_{q} \leq 1$ , $\int_{X} H (x) g (x) d μ (x) = \int_{Y} (\int_{X} ∣ F (x, y) ∣ g (x) d μ (x)) d ν (y)$ by [L2], and then [L3] yields $\int_{X} ∣ F (x, y) ∣ g (x) d μ (x) \leq ∥ F (\cdot, y) ∥_{p} ∥ g ∥_{q} \leq ∥ F (\cdot, y) ∥_{p} .$ Hence $\int_{X} H g d μ \leq \int_{Y} ∥ F (\cdot, y) ∥_{p} d ν (y) = M .$

1.3givenconstruct

Choose measurable sets $X_{1} \subseteq X_{2} \subseteq \dots$ with [given, construct] $μ (X_{m}) < \infty$ and $⋃_{m} X_{m} = X$ , and define $H_{m} : = \min (H, m) 1_{X_{m}} .$ Then $0 \leq H_{m} ↑ H$ pointwise, and each $H_{m}$ lies in $L^{p} (X)$ because $H_{m} \leq m 1_{X_{m}}$ .

2.1L1step 1.2step 1.3

Applying [L1] to each nonnegative $H_{m}$ and using $0 \leq H_{m} \leq H$ with [L1, step 1.2, step 1.3] step 1.2 gives $∥ H_{m} ∥_{p} \leq M$ for every $m$ .

3.1L4step 1.1step 2.1∎

By [L4], $\int_{X} H_{m} (x)^{p} d μ (x) ↑ \int_{X} H (x)^{p} d μ (x) .$ Since $\int H_{m}^{p} d μ = ∥ H_{m} ∥_{p}^{p} \leq M^{p}$ for every $m$ , the limit is finite and satisfies $∥ H ∥_{p}^{p} \leq M^{p}$ . Hence $H \in L^{p} (X)$ and $∥ H ∥_{p} \leq M = \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$ Together with step 1.1, this proves the theorem for all $1 \leq p < \infty$ .

Depends on

Used by

Dependency tree · two levels

26 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

John K. Hunter, Measure Theory (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)