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

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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.
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$ℓ^{p}$ includes into $ℓ^{r}$ for $p < r$

Statement

Let $1 \leq p < r \leq \infty$ . If $a = (a_{n})_{n \geq 0} \in ℓ^{p}$ , then $a \in ℓ^{r}$ . When $r < \infty$ one has

$∥ a ∥_{r} \leq ∥ a ∥_{p},$

and when $r = \infty$ one has

$∥ a ∥_{\infty} \leq ∥ a ∥_{p} .$

Facts & Assumptions

Given: A real sequence $a = (a_{n})_{n \geq 0}$ in $ℓ^{p}$ .

[L1]

$ℓ^{p}$ is $L^{p}$ of counting measure on $N$ ( $ℓ^{p}$ is the $L^{p}$ space of counting measure, Counting measure on an arbitrary set, Counting measure is a measure).

[L2]

Real powers obey the usual laws, and for fixed base $b > 1$ the map $x \mapsto b^{x}$ is strictly increasing (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential function is strictly increasing).

Proof

technique · For counting measure, only finitely many terms can exceed $1$ when a sequence lies in $\ell^p$. Split the series at that finite set and compare $|a_n|^r$ to $|a_n|^p$ on the tail where $|a_n|\le1$

1.1L1

If $r = \infty$ , then for each $n$ , [L1] $∣ a_{n} ∣^{p} \leq \sum_{k = 0}^{\infty} ∣ a_{k} ∣^{p} = ∥ a ∥_{p}^{p},$ so taking $p$ -th roots gives $∣ a_{n} ∣ \leq ∥ a ∥_{p}$ . Hence $∥ a ∥_{\infty} \leq ∥ a ∥_{p}$ .

1.2L1L2

Assume $r < \infty$ . Because $\sum_{n} ∣ a_{n} ∣^{p} < \infty$ , only finitely many indices can satisfy $∣ a_{n} ∣ > 1$ ; otherwise the $p$ -series would dominate the divergent sum of infinitely many $1$ 's. Thus $∣ a_{n} ∣ \leq 1$ for all sufficiently large $n$ . Since $r > p$ , one has $∣ a_{n} ∣^{r} \leq ∣ a_{n} ∣^{p}$ on that tail, so $\sum_{n} ∣ a_{n} ∣^{r}$ converges. [L1, L2].

2.1step 1.1step 1.2L1algebra

For every $n$ , $\sum_{k < n} ∣ a_{k} ∣^{r} \leq (\sup_{k} ∣ a_{k} ∣^{r - p}) \sum_{k < n} ∣ a_{k} ∣^{p} \leq ∥ a ∥_{\infty}^{r - p} \sum_{k < n} ∣ a_{k} ∣^{p} .$ Using step 1.1, this becomes $\sum_{k < n} ∣ a_{k} ∣^{r} \leq ∥ a ∥_{p}^{r - p} \sum_{k < n} ∣ a_{k} ∣^{p} .$ Letting $n \to \infty$ yields $∥ a ∥_{r}^{r} \leq ∥ a ∥_{p}^{r}$ , hence $∥ a ∥_{r} \leq ∥ a ∥_{p}$ .

3.1step 1.1step 2.1∎

Step 1.1 proves the endpoint $r = \infty$ , and step 2.1 proves the finite- $r$ estimate. [step 1.1, step 2.1].

Depends on

Used by

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Dependency tree · two levels

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Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorems 8.12 and 8.13 (standard reference, not scraped)
John K. Hunter, Measure Theory, Chapter 17 overview (standard reference, not scraped)

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