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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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].

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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