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A function whose trace is at most a level has positive part in the zero-boundary space

Statement

Assume Countable Choice and the Axiom of Choice, inherited through the published trace and density suppliers named below. Let $n \geq 2$ , let $Ω \subset R^{n}$ be a bounded $C^{1}$ domain (Bounded C^k domains and boundary charts), let $u \in H^{1} (Ω; R)$ , and let $T$ be the trace operator of The $L^{p}$ trace operator on a bounded $C^{1}$ domain. Then $T u \leq 0$ a.e. on $\partial Ω$ implies $u^{+} \in H_{0}^{1} (Ω)$ , and conversely $u^{+} \in H_{0}^{1} (Ω)$ implies $T u \leq 0$ a.e.; more generally, for every $k \in R$ , $(u - k)^{+} \in H_{0}^{1} (Ω)$ if and only if $T u \leq k$ a.e. on $\partial Ω$ . In particular the weak boundary order of Weak subsolutions and supersolutions of a divergence-form equation is the pointwise trace order, and the two conventions give the same boundary supremum $\sup_{\partial Ω} u = {ess sup}_{\partial Ω} T u$ (with value $+ \infty$ only if the trace is not essentially bounded above).

Facts & Assumptions

Given: Countable Choice and the Axiom of Choice; a bounded $C^{1}$ domain $Ω \subset R^{n}$ , $n \geq 2$ ; a real class $u \in H^{1} (Ω; R)$ ; the trace $T$ ; and a real level $k$ .

[F1]

$T : W^{1, 2} (Ω; R) \to L^{2} (\partial Ω; R)$ is linear and bounded, and $T v = v ∣_{\partial Ω}$ for every $v \in C (\overline{Ω}) \cap H^{1} (Ω)$ (The $L^{p}$ trace operator on a bounded $C^{1}$ domain, The trace agrees with classical restriction for continuous Sobolev functions).

[F2]

Assume the Axiom of Choice. Smooth functions on $\overline{Ω}$ (restrictions of $C_{c}^{\infty} (R^{n})$ functions) are dense in $H^{1} (Ω)$ , and $C_{c}^{\infty} (Ω)$ is dense in $H_{0}^{1} (Ω)$ by definition (Ambient smooth restrictions are dense on bounded C^k domains, Zero-boundary Sobolev space as a norm closure).

[F3]

Assume the Axiom of Choice. ${w \in H^{1} (Ω) : T w = 0} = H_{0}^{1} (Ω)$ (The kernel of the trace is the closure of the test functions).

[F4]

Assume the Axiom of Choice. If $w, w_{j} \in H^{1} (Ω; R)$ with $w_{j} \to w$ in $H^{1}$ , then $w_{j}^{+} \to w^{+}$ in $H^{1}$ : pointwise $∣ w_{j}^{+} - w^{+} ∣ \leq ∣ w_{j} - w ∣$ and $D (w_{j}^{+} - w^{+}) = 1_{{w_{j} > 0}} D w_{j} - 1_{{w > 0}} D w = 1_{{w_{j} > 0}} D (w_{j} - w) + (1_{{w_{j} > 0}} - 1_{{w > 0}}) D w$ , whose first term tends to $0$ in $L^{2}$ . Every subsequence has a further subsequence with $w_{j} \to w$ a.e.: choose the further terms with $∥ w_{j} - w ∥_{2}^{2} \leq 2^{- 3 j}$ , so $∣ {∣ w_{j} - w ∣ > 2^{- j}} ∣ \leq 2^{- j}$ and countable subadditivity makes the limsup null. Along this further subsequence the indicator difference tends to zero where $w \neq 0$ , while $D w = 0$ a.e. where $w = 0$ ; dominated convergence with $4 ∣ D w ∣^{2}$ makes the second term tend to zero in $L^{2}$ . If the full positive-part sequence did not converge, a subsequence with errors bounded below would contradict this argument. Therefore $w_{j}^{+} \to w^{+}$ in $H^{1}$ (Positive-part truncation calculus and admissible cut-off weak tests, Positive, negative, and truncated Sobolev functions).

[F5]

Weak boundary order: $u \leq k$ on $\partial Ω$ means $(u - k)^{+} \in H_{0}^{1} (Ω)$ , and $\sup_{\partial Ω} u = \inf {k : u \leq k on \partial Ω}$ with $\inf \emptyset = + \infty$ (Weak subsolutions and supersolutions of a divergence-form equation).

Proof

technique · direct; approximate by functions smooth up to the boundary, where the trace is the boundary restriction and commutes with truncation, then pass to the limit

1.1givenF1F2

Fix $k \in R$ and put $w : = u - k \in H^{1} (Ω; R)$ and $w_{k} : = (u - k)^{+}$ . By [F2] choose $w_{j} \in C^{\infty} (\overline{Ω})$ with $w_{j} \to w$ in $H^{1}$ . For each $j$ , $w_{j}^{+}$ is continuous on $\overline{Ω}$ as the maximum of the continuous functions $w_{j}$ and $0$ , and it lies in $H^{1} (Ω)$ by the Lipschitz chain rule, so [F1] gives $T w_{j}^{+} = w_{j}^{+} ∣_{\partial Ω} = (w_{j} ∣_{\partial Ω})^{+} = (T w_{j})^{+}$ (the last equality using $T w_{j} = w_{j} ∣_{\partial Ω}$ from [F1]); moreover $T w_{j} \to T w$ in $L^{2} (\partial Ω)$ , so $(T w_{j})^{+} \to (T w)^{+}$ in $L^{2} (\partial Ω)$ because $t \mapsto t^{+}$ is $1$ -Lipschitz on $R$ .

2.1step 1.1F1F4

By [F4], $w_{j}^{+} \to w^{+}$ in $H^{1} (Ω)$ , so the continuity of $T$ in [F1] gives $T w_{j}^{+} \to T w^{+}$ in $L^{2} (\partial Ω)$ . Since step 1.1 gives $T w_{j}^{+} = (T w_{j})^{+} \to (T w)^{+}$ in the same space, uniqueness of $L^{2}$ limits yields $T (w^{+}) = (T w)^{+}$ a.e. on $\partial Ω$ .

3.1step 2.1F3algebra

Consequently, by [F3], $w^{+} \in H_{0}^{1} (Ω) ⟺ T w^{+} = 0$ in $L^{2} (\partial Ω) ⟺ (T w)^{+} = 0$ a.e. $⟺ T w \leq 0$ a.e. on $\partial Ω$ ; since $w = u - k$ and $w^{+} = (u - k)^{+}$ , this is the asserted equivalence $(u - k)^{+} \in H_{0}^{1} (Ω) ⟺ T u \leq k$ a.e.

4.1step 3.1F3F5∎

The boundary supremum. By [F5] and step 3.1, the admissible levels are ${k \in R : u \leq k$ on $\partial Ω} = {k \in R : T u \leq k$ a.e. $} = [{ess sup}_{\partial Ω} T u, + \infty)$ when the trace is essentially bounded above, and the empty set when it is not; the infimum is therefore ${ess sup}_{\partial Ω} T u$ in the first case and $+ \infty$ in the second, which proves the boundary-supremum identification. The argument uses only the declared Countable Choice and Axiom of Choice.

A function whose trace is at most a level has positive part in the zero-boundary space

Statement

Facts & Assumptions

Proof

Depends on

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Sources