TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < \infty$ .

If $1 \leq p < r < \infty$ and $f \in L^{r} (μ)$ , then $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p - 1 / r} ∥ f ∥_{r} .$
If $1 \leq p < \infty$ and $f \in L^{\infty} (μ)$ , then $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p} ∥ f ∥_{\infty} .$

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ .

[L1]

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

[L2]

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

[L3]

$L^{r} (μ)$ and $L^{\infty} (μ)$ are the measurable-function spaces of The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L4]

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

Proof

technique · Write $|f|^p$ as $|f|^p\cdot1$ and apply Holder with exponents $r/p$ and $r/(r-p)$. The finite total measure contributes the factor $\mu(X)^{1/p-1/r}$

1.1L1L2L3givenalgebra

Suppose $1 \leq p < r < \infty$ and put $a : = \frac{r}{p}, b : = \frac{r}{r - p} .$ Then $a, b \in (1, \infty)$ and $1 / a + 1 / b = 1$ , so [L2] makes them conjugate. Apply [L1] to the functions $∣ f ∣^{p}$ and $1$ : $\int ∣ f ∣^{p} d μ \leq {(\int ∣ f ∣^{p a} d μ)}^{1 / a} {(\int 1^{b} d μ)}^{1 / b} = {(\int ∣ f ∣^{r} d μ)}^{p / r} μ (X)^{1 - p / r} .$ Taking $p$ -th roots gives the claimed bound.

1.2L3L4givenalgebra

If $f \in L^{\infty} (μ)$ , then $\int ∣ f ∣^{p} d μ \leq ∥ f ∥_{\infty}^{p} μ (X) .$ Indeed, $∣ f ∣^{p} \leq ∥ f ∥_{\infty}^{p}$ almost everywhere by [L4]. Thus $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p} ∥ f ∥_{\infty}$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the finite-measure inclusion laws. [step 1.1, step 1.2].

Depends on

Used by

Dependency tree · two levels

17 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

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem 8.2 (standard reference, not scraped)
John K. Hunter, Measure Theory, Chapter 17 overview (standard reference, not scraped)

Finite-measure LrLr includes into LpLp for p<rp<r