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Minkowski's inequality for integrals, including $p = \infty$

Statement

Let $(X, A, μ)$ be a measure space.

If $1 \leq p < \infty$ and $f, g \in L^{p} (μ)$ , then $∥ f + g ∥_{p} \leq ∥ f ∥_{p} + ∥ g ∥_{p} .$
If $f, g \in L^{\infty} (μ)$ , then $∥ f + g ∥_{\infty} \leq ∥ f ∥_{\infty} + ∥ g ∥_{\infty} .$

Facts & Assumptions

Given: A measure space and functions $f, g$ in the spaces named in the relevant clause of the Statement.

[L1]

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

[L2]

Membership in $L^{p} (μ)$ and $L^{\infty} (μ)$ is defined in The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L3]

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

[L4]

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

[L5]

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

[L6]

Holder's inequality for finite sums gives, for nonnegative reals $a, b$ , $a + b \leq 2^{1 / q} (a^{p} + b^{p})^{1 / p}$ when $q = p / (p - 1)$ (Holder's inequality for finite sums and conjugate real exponents).

Proof

technique · For $1 < p < infinity$, first use the two-term finite Holder inequality to show $|f + g|^p$ is integrable, then apply the standard Holder step to $|f + g| |f + g|^{p - 1}$. The cases $p = 1$ and $p = \infty$ are handled directly from subadditivity of absolute value and the essential-supremum bound

1.1L4L5given

If $p = 1$ , then $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣$ pointwise, so [L4, L5, given] $∥ f + g ∥_{1} = \int ∣ f + g ∣ d μ \leq \int ∣ f ∣ d μ + \int ∣ g ∣ d μ = ∥ f ∥_{1} + ∥ g ∥_{1} .$

1.2L1L2L4L5L6givenalgebra

Assume $1 < p < \infty$ and let $q : = p / (p - 1)$ . Then [L1, L2, L4, L5, L6, given, algebra] $\int ∣ f + g ∣^{p} d μ \leq 2^{p - 1} (\int ∣ f ∣^{p} d μ + \int ∣ g ∣^{p} d μ) < \infty .$ Indeed, [L6] applied pointwise to the two-term families $(∣ f (x) ∣, ∣ g (x) ∣)$ and $(1, 1)$ gives $∣ f (x) ∣ + ∣ g (x) ∣ \leq 2^{1 / q} (∣ f (x) ∣^{p} + ∣ g (x) ∣^{p})^{1 / p},$ so $∣ f + g ∣^{p} \leq (∣ f ∣ + ∣ g ∣)^{p} \leq 2^{p - 1} (∣ f ∣^{p} + ∣ g ∣^{p})$ pointwise. Thus $f + g \in L^{p} (μ)$ . Put $C : = ∥ f + g ∥_{p}$ . If $C = 0$ , the claim is immediate. Otherwise $∣ f + g ∣^{p} = ∣ f + g ∣ ∣ f + g ∣^{p - 1} \leq ∣ f ∣ ∣ f + g ∣^{p - 1} + ∣ g ∣ ∣ f + g ∣^{p - 1} .$ Because $(p - 1) q = p$ , the function $∣ f + g ∣^{p - 1}$ lies in $L^{q} (μ)$ and has $q$ -norm $C^{p - 1}$ . Integrating and applying [L1] with conjugate exponents $p$ and $q$ to each term yields $C^{p} \leq ∥ f ∥_{p} {(\int ∣ f + g ∣^{(p - 1) q} d μ)}^{1 / q} + ∥ g ∥_{p} {(\int ∣ f + g ∣^{(p - 1) q} d μ)}^{1 / q} .$ Since $(p - 1) q = p$ , this becomes $C^{p} \leq (∥ f ∥_{p} + ∥ g ∥_{p}) C^{p - 1} .$ If $C > 0$ , divide by $C^{p - 1}$ to obtain the claim.

1.3L2L3given

For $p = \infty$ , let $M : = ∥ f ∥_{\infty}$ and $N : = ∥ g ∥_{\infty}$ . Then [L2, L3, given] $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣ \leq M + N .$ Indeed, outside the union of the two null exceptional sets supplied by [L3], one has $∣ f ∣ \leq M$ and $∣ g ∣ \leq N$ . Therefore $∥ f + g ∥_{\infty} \leq M + N$ .

2.1step 1.1step 1.2step 1.3∎

Steps 1.1, 1.2, and 1.3 prove the $p = 1$ , $1 < p < \infty$ , and $p = \infty$ cases.

Depends on

Used by

Dependency tree · two levels

29 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, Minkowski's Inequality (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 7.5 (standard reference, not scraped)

Minkowski's inequality for integrals, including p=∞p=∞