LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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.

A dyadic summability estimate for decreasing level-set sequences

Statement

Let $d \geq 1$ , $0 < θ < 1$ , $1 \leq p < \infty$ with $p θ < d$ and $T > 1$ . Let $(a_{k})_{k \in Z}$ be a bounded nonnegative nonincreasing sequence of real numbers with $a_{k} = 0$ for all sufficiently large $k$ . Then $\sum_{k \in Z} a_{k}^{(d - p θ) / d} T^{k} \leq C (d, p, θ, T) \sum_{k \in Z : a_{k} \neq 0} a_{k + 1} a_{k}^{- p θ / d} T^{k} .$ The constant is explicit: $C = T^{d / (d - p θ)}$ . The argument uses no choice principle; all sums are series of nonnegative terms.

Facts & Assumptions

Given: integers $d \geq 1$ and $k \in Z$ indices, numbers $0 < θ < 1$ , $1 \leq p < \infty$ with $p θ < d$ , a real $T > 1$ , and a bounded nonnegative nonincreasing sequence $(a_{k})_{k \in Z}$ with $a_{k} = 0$ for all sufficiently large $k$ . Put $t : = p θ / d \in (0, 1)$ , $α : = 1 / t > 1$ and $β : = 1 / (1 - t) > 1$ , so that $1 / α + 1 / β = 1$ .

[F1]

Hölder's inequality. On a measure space $(X, A, μ)$ , for conjugate exponents $α, β \in (1, \infty)$ and nonnegative measurable $f, g$ with finite respective norms, $\int f g d μ \leq (\int f^{α} d μ)^{1 / α} (\int g^{β} d μ)^{1 / β}$ , which is the finite-norm form used below. Step 1.1 establishes the required finite sums before the application. (Holder's inequality for integrals, including the endpoint cases)

Proof

technique · Shift the index, factor each term of the shifted sum into a product whose two factors have the two critical exponents, apply Hölder's inequality on the counting measure, and solve the resulting inequality for the unknown sum

1.1givenalgebra

Since $a_{k} = 0$ for all $k \geq N$ and $(a_{k})$ is bounded by some $M \geq 0$ , the sum $A : = \sum_{k \in Z} a_{k}^{(d - p θ) / d} T^{k}$ satisfies $A \leq M^{(d - p θ) / d} \sum_{k < N} T^{k} < \infty$ , and the sum $B : = \sum_{k : a_{k} \neq 0} a_{k + 1} a_{k}^{- p θ / d} T^{k}$ satisfies $0 \leq B \leq \sum_{k : a_{k} \neq 0} a_{k}^{(d - p θ) / d} T^{k} = A < \infty$ because $a_{k + 1} \leq a_{k}$ with $a_{k} > 0$ implies $a_{k + 1} a_{k}^{- p θ / d} \leq a_{k}^{1 - p θ / d} = a_{k}^{(d - p θ) / d}$ . Also $a_{k} = 0$ implies $a_{k + 1} \leq a_{k} = 0$ , so $a_{k + 1} = 0$ .

2.1F1step 1.1algebra

Shifting the index in step 1.1 and dropping exactly the vanishing terms, $\frac{1}{T} A = \sum_{k \in Z} a_{k + 1}^{(d - p θ) / d} T^{k} = \sum_{k : a_{k} \neq 0} a_{k + 1}^{(d - p θ) / d} T^{k}$ . For each $k$ with $a_{k} \neq 0$ the factorization $a_{k + 1}^{1 - t} T^{k} = (a_{k}^{t / β} T^{k / α}) (a_{k + 1}^{1 / β} a_{k}^{- t / β} T^{k / β})$ holds, because $t / β - t / β = 0$ , $1 / β = 1 - t$ and $1 / α + 1 / β = 1$ . Applying [F1] with the counting measure on the set ${k : a_{k} \neq 0}$ to these two factors gives $\frac{1}{T} A \leq (\sum_{k} a_{k}^{1 - t} T^{k})^{t} (\sum_{k : a_{k} \neq 0} a_{k + 1} a_{k}^{- t} T^{k})^{1 - t} = A^{t} B^{1 - t}$ , since raising the first factor-sum to the power $α = 1 / t$ returns $\sum_{k} a_{k}^{1 - t} T^{k}$ and raising the second to $β = 1 / (1 - t)$ returns $B$ .

3.1step 1.1step 2.1algebra∎

If $A = 0$ then every $a_{k} = 0$ and both sides of the asserted inequality are $0$ . Otherwise $0 < A < \infty$ by step 1.1, so dividing step 2.1 by $T A^{t}$ gives $(1 / T) A^{1 - t} \leq B^{1 - t}$ , hence $B^{1 - t} \geq (1 / T) A^{1 - t} > 0$ and therefore $A \leq T^{1 / (1 - t)} B$ . Substituting $t = p θ / d$ gives $1 / (1 - t) = d / (d - p θ)$ and $A \leq T^{d / (d - p θ)} B$ , which is the assertion with $C = T^{d / (d - p θ)}$ . No choice principle is used: both series are sums of nonnegative real terms over a countable index set, evaluated as suprema of finite partial sums.

Depends on

Holder's inequality for integrals, including the endpoint cases

Used by

The critical fractional Sobolev inequality on ℝᵈ Theorem

Dependency tree · two levels

11 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

Eleonora Di Nezza, Giampiero Palatucci and Enrico Valdinoci, Hitchhiker's guide to the fractional Sobolev spaces (arXiv:1104.4345, survey) (standard reference, not scraped)