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Whole-space inequalities transfer through a Sobolev extension

Statement

Assume the Axiom of Choice. Let $k \in N_{0}$ , $1 \leq p \leq \infty$ , $K \in {R, C}$ , and let $Ω \subseteq R^{n}$ be open. Let $E : W^{k, p} (Ω; K) ⟶ W^{k, p} (R^{n}; K)$ be a bounded linear extension operator, so that $(E u) ∣_{Ω} = u$ almost everywhere on $Ω$ for every class $u$ , and let $∥ E ∥$ be its operator norm. Suppose a whole-space functional $N$ on Sobolev classes, together with its restrictions $N_{Ω}$ to the classes of $Ω$ , satisfies $N_{Ω} (F ∣_{Ω}) \leq N (F) and N (F) \leq C ∥ F ∥_{W^{k, p} (R^{n})}$ for every $F \in W^{k, p} (R^{n}; K)$ and a constant $C$ independent of $F$ . Then $N_{Ω} (u) \leq C ∥ E ∥ ∥ u ∥_{W^{k, p} (Ω)} for every u \in W^{k, p} (Ω; K) .$ In particular, if $1 \leq q \leq \infty$ and a whole-space Sobolev inequality $∥ F ∥_{L^{q} (R^{n})} \leq C ∥ F ∥_{W^{k, p} (R^{n})}$ is available, then $∥ u ∥_{L^{q} (Ω)} \leq C ∥ E ∥ ∥ u ∥_{W^{k, p} (Ω)}$ for every $u$ . The corollary is conditional on that whole-space inequality and asserts no embedding theorem itself; every bounded $C^{k}$ domain supplies an admissible operator $E$ through Bounded C^k domains admit integer-order Sobolev extension.

Facts & Assumptions

Given: the Axiom of Choice; $k \in N_{0}$ ; $1 \leq p \leq \infty$ ; $K \in {R, C}$ ; an open $Ω \subseteq R^{n}$ ; a bounded linear extension operator $E$ with right-inverse property and norm $∥ E ∥$ ; a functional $N$ with restrictions $N_{Ω}$ satisfying the two displayed hypotheses with constant $C$ ; and a class $u \in W^{k, p} (Ω; K)$ .

[F1]

Extension operator: $E : W^{k, p} (Ω; K) \to W^{k, p} (R^{n}; K)$ is bounded and linear with $(E u) ∣_{Ω} = u$ as an almost-everywhere class on $Ω$ for every $u$ , and its operator norm is $∥ E ∥ = \sup {∥ E u ∥ : u \in W^{k, p} (Ω; K), ∥ u ∥_{W^{k, p} (Ω)} \leq 1}$ (Sobolev extension domains and extension operators).

[F2]

Restriction is a well-defined operation on Sobolev classes: $F ∣_{Ω} \in W^{k, p} (Ω; K)$ with $D^{α} (F ∣_{Ω}) = (D^{α} F) ∣_{Ω}$ almost everywhere, and it is a contraction (Bounded restriction and cutoff localisation in Sobolev spaces).

[F3]

Restriction monotonicity of $N$ : $N_{Ω} (F ∣_{Ω}) \leq N (F)$ for every $F \in W^{k, p} (R^{n}; K)$ , and the whole-space bound $N (F) \leq C ∥ F ∥_{W^{k, p} (R^{n})}$ , both hypotheses of the statement.

[F4]

The Sobolev norm is the finite derivative sum of Integer-order Sobolev spaces and their norms, and $∥ E u ∥ \leq ∥ E ∥ ∥ u ∥$ holds for every $u$ by the definition of the operator norm in [F1].

[F5]

Bounded $C^{k}$ domains: for $k \geq 1$ and every $1 \leq p \leq \infty$ there is a bounded linear extension operator $W^{k, p} (Ω; K) \to W^{k, p} (R^{n}; K)$ for every bounded $C^{k}$ domain in the graph sense, and extension by zero supplies the case $k = 0$ on any open set (Bounded C^k domains admit integer-order Sobolev extension).

Choice use. The Axiom of Choice enters only through the published interfaces of [F2] and [F5], which invoke the Countable Choice they require; [F5] also invokes it for the chart and partition-of-unity steps of the extension construction. The three-line norm chain of the proof itself uses no choice.

Proof

technique · direct

1.1F1given

Fix $u \in W^{k, p} (Ω; K)$ and put $F : = E u \in W^{k, p} (R^{n}; K)$ . By the right-inverse property of [F1], $F ∣_{Ω} = u$ as an almost-everywhere class on $Ω$ .

2.1F3step 1.1

Apply the restriction monotonicity of [F3] to the pair $(F, Ω)$ : $N_{Ω} (u) = N_{Ω} (F ∣_{Ω}) \leq N (F) = N (E u)$ .

2.2F3step 1.1

Apply the whole-space bound of [F3] to $F = E u$ : $N (E u) \leq C ∥ E u ∥_{W^{k, p} (R^{n})}$ .

2.3F1F4step 1.1

Apply the operator-norm inequality of [F4] to $u$ : $∥ E u ∥_{W^{k, p} (R^{n})} \leq ∥ E ∥ ∥ u ∥_{W^{k, p} (Ω)}$ .

3.1step 2.1step 2.2step 2.3

Chaining steps 2.1, 2.2 and 2.3 gives $N_{Ω} (u) \leq C ∥ E ∥ ∥ u ∥_{W^{k, p} (Ω)}$ for the fixed class $u$ ; since $u$ was arbitrary, the first assertion holds.

4.1F2F3step 3.1given

$L^{q}$ instance. Let $1 \leq q \leq \infty$ and set $N (F) : = ∥ F ∥_{L^{q} (R^{n})}$ and $N_{Ω} (v) : = ∥ v ∥_{L^{q} (Ω)}$ , with the value $+ \infty$ when the class is not in $L^{q}$ . The inclusion $Ω \subseteq R^{n}$ gives $\int_{Ω} ∣ F ∣^{q} \leq \int_{R^{n}} ∣ F ∣^{q}$ for $q < \infty$ and ${ess sup}_{Ω} ∣ F ∣ \leq {ess sup}_{R^{n}} ∣ F ∣$ for $q = \infty$ ; hence $N_{Ω} (F ∣_{Ω}) \leq N (F)$ , where the left side is interpreted through the class of [F2]. If the whole-space inequality $∥ F ∥_{L^{q} (R^{n})} \leq C ∥ F ∥_{W^{k, p} (R^{n})}$ is available, the remaining hypothesis of [F3] holds with that same constant, and step 3.1 yields $∥ u ∥_{L^{q} (Ω)} \leq C ∥ E ∥ ∥ u ∥_{W^{k, p} (Ω)}$ .

5.1F5step 4.1∎

Domains. If $Ω$ is a bounded $C^{k}$ domain with $k \geq 1$ , [F5] supplies an admissible operator $E$ for every $1 \leq p \leq \infty$ , so step 4.1 transfers any available whole-space $L^{q}$ inequality to $Ω$ with the extension constant of that operator; for $k = 0$ extension by zero supplies the analogous operator on any open set. No embedding is proved here: the implication is conditional on the whole-space inequality, and the conclusion is stated only for the functional $N$ and the operator $E$ that are given.

Whole-space inequalities transfer through a Sobolev extension

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources