TheoremStatement: AI-adaptedProof: 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.

The p=1 Gagliardo-Nirenberg-Sobolev inequality

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 2$ . There is a constant $C (n)$ such that $∥ u ∥_{L^{n / (n - 1)} (R^{n})} \leq C (n) ∥ D u ∥_{L^{1} (R^{n})}$ for every $u \in C_{c}^{\infty} (R^{n}; K)$ .

Facts & Assumptions

Given: Countable Choice; an integer $n \geq 2$ ; a field $K \in {R, C}$ ; and a function $u \in C_{c}^{\infty} (R^{n}; K)$ .

[F1]

Vector-valued fundamental theorem: if $f$ is differentiable with integrable derivative on an interval, then $\int_{a}^{b} f^{'} = f (b) - f (a)$ (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz).

[F2]

Tonelli's theorem on sigma-finite products: iterated integrals of nonnegative product-measurable functions may be computed in any order and partial integrals may be renamed (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F3]

Holder's inequality: if $1 / r = 1 / p + 1 / q$ and the indicated spaces are over one measure space, then $∥ f g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ (Generalized Holder inequality puts products into $L^{r}$ ).

[F4]

$L^{p}$ is the quotient by almost-everywhere null functions, and complex-valued Lebesgue spaces use the componentwise conventions (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions).

[F5]

Countable Choice, assumed for the measure-theoretic interfaces above (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1F1F4givenalgebra

Sections and pointwise bounds. Fix $j \in {1, \dots, n}$ and write ${\hat{x}}_{j}$ for the coordinates other than $x_{j}$ . For each fixed ${\hat{x}}_{j}$ , the section $t \mapsto u (x_{1}, \dots, t, \dots, x_{n})$ is smooth and compactly supported, so the fundamental theorem [F1] applied on an interval containing the support and the estimate $∣ \partial_{j} u ∣ \leq ∣ D u ∣$ give $∣ u (x) ∣ \leq \int_{R} ∣ D u (x_{1}, \dots, t, \dots, x_{n}) ∣ d t = : P_{j} ({\hat{x}}_{j})$ for every $x$ .

1.2F2F3algebra

The product-integral lemma. For $d \geq 2$ and nonnegative integrable functions $f_{j}$ on $R^{d - 1}$ , each independent of the $j$ -th coordinate, one has $\int_{R^{d}} \prod_{j = 1}^{d} f_{j} ({\hat{x}}_{j})^{1 / (d - 1)} d x \leq \prod_{j = 1}^{d} (\int f_{j})^{1 / (d - 1)}$ . For $d = 2$ this is Tonelli [F2]. For $d > 2$ , integrate first in $x_{d}$ and apply [F3] with $d - 1$ equal exponents to the factors $j < d$ . Put $g_{j} : = \int_{R} f_{j} d x_{d}$ for $j < d$ ; the resulting upper bound is $\int_{R^{d - 1}} f_{d}^{1 / (d - 1)} \prod_{j < d} g_{j}^{1 / (d - 1)}$ . Holder with exponents $d - 1$ and $(d - 1) / (d - 2)$ bounds this by $(\int f_{d})^{1 / (d - 1)} (\int \prod_{j < d} g_{j}^{1 / (d - 2)})^{(d - 2) / (d - 1)}$ . The induction hypothesis in dimension $d - 1$ , followed by Tonelli, gives the required product of the $\int f_{j}$ . Zero integrals make the integrand zero almost everywhere, so they cause no division.

2.1F5step 1.1algebra

Product and root. Multiplying the $n$ pointwise inequalities of step 1.1 and taking the $(n - 1)$ -th root gives $∣ u (x) ∣^{n / (n - 1)} \leq \prod_{j = 1}^{n} P_{j} ({\hat{x}}_{j})^{1 / (n - 1)}$ for every $x$ .

3.1F2F4step 2.1step 1.2algebra∎

Apply step 1.2 with $d = n$ and $f_{j} = P_{j}$ . Tonelli gives $\int_{R^{n - 1}} P_{j} = \int_{R^{n}} ∣ D u ∣ = ∥ D u ∥_{1} < \infty$ . Step 2.1 therefore yields $\int ∣ u ∣^{n / (n - 1)} \leq ∥ D u ∥_{1}^{n / (n - 1)}$ . Taking the $(n - 1) / n$ -th power proves the assertion with $C (n) = 1$ .

Source notes

Kinnunen's Theorem 3.3 computes the product of the $n$ one-dimensional primitive estimates and integrates one variable at a time with the generalized Holder inequality for $(n - 1)$ factors; the proof above records a dimension induction for the product-integral inequality. The constant obtained is $1$ , which is not sharp but is dimension-only as asserted. The argument is the case $p = 1$ separated in the plan because the power-and-Holder reduction used for $1 < p < n$ is unavailable at the endpoint.

The p=1 Gagliardo-Nirenberg-Sobolev inequality

Statement

Facts & Assumptions

Proof

Source notes

Depends on

Dependency tree · 2 levels

Sources