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

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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

Proof technique: For counting measure, only finitely many terms can exceed $1$ when a sequence lies in $ℓ^{p}$ . Split the series at that finite set and compare $∣ a_{n} ∣^{r}$ to $∣ a_{n} ∣^{p}$ on the tail where $∣ a_{n} ∣ \leq 1$ .

1.1

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 $∥ a ∥_{\infty} \leq ∥ a ∥_{p}$ .

1.2

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.

L1L2

2.1

For every $n$ , [step 1.1, step 1.2, L1, algebra] $\sum_{k < n} ∣ a_{k} ∣^{r} \leq (\sup_{k} ∣ a_{k} ∣^{r - p}) \sum_{k < n} ∣ a_{k} ∣^{p} \leq ∥ a ∥_{\infty}^{r - p} \sum_{k < n} ∣ a_{k} ∣^{p} .$ 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 $∥ a ∥_{r} \leq ∥ a ∥_{p}$ .

3.1

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

step 1.1step 2.1∎

Depends on

Used by

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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)

ℓpℓp includes into ℓrℓr for p<rp<r