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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.

Depends on

Used by

Dependency tree · two levels

59 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

Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete 158-page graduate notes) (standard reference, not scraped)