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BV Sobolev inequality in full space

Statement

Assume Countable Choice and let $n \geq 2$ . Every $u \in B V (R^{n})$ satisfies $∥ u ∥_{L^{n / (n - 1)} (R^{n})} \leq T V (u; R^{n}) .$

Statement references

Spatial BV and total variation.

The Lp spaces are defined using Lebesgue measure on Rn.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , and $u \in B V (R^{n})$ .

[A1]

Countable Choice is needed for the strict smooth approximation used below. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Every full-space BV function has a compactly supported smooth strict approximation converging in $L^{1}$ and in gradient integral. (Strict smooth approximation in BV)

[F2]

The Loomis–Whitney integral inequality bounds the coordinate-product integral by the product of its $(n - 1)$ -dimensional integrals. (Loomis–Whitney integral inequality)

[F3]

For a smooth function with continuous derivative, the one-dimensional fundamental theorem gives its value from the integral of its derivative along each coordinate line. (Fundamental theorem of calculus for Banach-valued continuous curves)

[F4]

Tonelli allows the coordinate-line integrals to be assembled into the full-space integrals below. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

Given: The hypotheses listed above.

1.1F3F4givenalgebra

First let $f \in C_{c}^{\infty} (R^{n})$ . For each $i$ , put $g_{i} (x_{\hat{i}}) : = \int_{R} ∣ \partial_{i} f (x_{1}, \dots, x_{i - 1}, s, x_{i + 1}, \dots, x_{n}) ∣ d s .$ Tonelli [F4] shows $g_{i}$ is nonnegative measurable and integrable. By [F3], integrating from a point to the left of the compact support up to $x_{i}$ gives $∣ f (x) ∣ \leq g_{i} (x_{\hat{i}})$ for every coordinate $i$ . Multiplication and raising to the power $1 / (n - 1)$ yield $∣ f (x) ∣^{n / (n - 1)} \leq \prod_{i = 1}^{n} g_{i} (x_{\hat{i}})^{1 / (n - 1)} .$

2.1F2step 1.1algebra

Integrating this inequality and applying [F2] gives $∥ f ∥_{L^{n / (n - 1)}} \leq \prod_{i = 1}^{n} ∥ \partial_{i} f ∥_{1}^{1 / n} \leq \frac{1}{n} \sum_{i = 1}^{n} ∥ \partial_{i} f ∥_{1} \leq \frac{1}{\sqrt{n}} \int_{R^{n}} ∣ \nabla f ∣ d x .$ The middle inequality is arithmetic-geometric mean; the last follows pointwise from $\sum_{i} ∣ \partial_{i} f ∣ \leq \sqrt{n} ∣ \nabla f ∣$ . Thus the desired inequality holds for smooth compactly supported $f$ , with the stronger constant $1 / \sqrt{n}$ .

3.1A1F1F4step 2.1choose∎

By [A1] and [F1], choose $f_{k} \in C_{c}^{\infty} (R^{n})$ with $f_{k} \to u$ in $L^{1}$ and $\int ∣ \nabla f_{k} ∣ \to T V (u; R^{n})$ . Select a subsequence with $∥ f_{k_{j}} - u ∥_{1} < 2^{- 2 j}$ ; Tonelli [F4] gives $\int \sum_{j} ∣ f_{k_{j}} - u ∣ < \infty$ , hence $f_{k_{j}} \to u$ almost everywhere. Fatou and step 2.1 imply $\int ∣ u ∣^{n / (n - 1)} d x \leq {lim inf}_{j} \int ∣ f_{k_{j}} ∣^{n / (n - 1)} d x \leq T V (u; R^{n})^{n / (n - 1)} .$ Taking the $(n - 1) / n$ power proves the stated inequality and at the same time shows $u \in L^{n / (n - 1)}$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Gláucio Terra, Introduction to Geometric Measure Theory — Lecture Notes, version 1.6 (standard reference, not scraped)
Piotr Rybka, The space in variational and evolution problems — Lecture Notes in Mathematical Sciences 15 (standard reference, not scraped)