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The obstacle admissible set is nonempty, convex, closed and weakly closed

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), inherited through the truncation and trace suppliers named below. Let $n \geq 1$ ; when $n = 1$ use a bounded interval and the endpoint trace of Endpoint trace commutes with Sobolev truncation on an interval, and when $n \geq 2$ use a bounded $C^{1}$ domain with the published trace definitions (Bounded C^k domains and boundary charts). Let $ψ \in H^{1} (Ω; R)$ satisfy $T ψ \leq 0$ (componentwise at the endpoints for $n = 1$ , almost everywhere on $\partial Ω$ for $n \geq 2$ ), so that $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ is the admissible set of The closed convex obstacle set and the obstacle variational inequality (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms). Then $K$ is nonempty, convex, closed in the norm of $H_{0}^{1} (Ω)$ and weakly sequentially closed: if $(v_{j}) \subseteq K$ and $v_{j} ⇀ v$ in $H_{0}^{1} (Ω)$ , then $v \in K$ (Weak convergence of nets and sequences).

Facts & Assumptions

Given: The setting above and an obstacle $ψ \in H^{1} (Ω; R)$ with $T ψ \leq 0$ ; the admissible set $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ .

[F1]

The closed convex obstacle set and the obstacle variational inequality: the boundary condition $T ψ \leq 0$ is equivalent to $ψ^{+} \in H_{0}^{1} (Ω)$ (through the interval lemma for $n = 1$ and A function whose trace is at most a level has positive part in the zero-boundary space for $n \geq 2$ ), and $K$ is defined by the representative-independent almost-everywhere inequality $v \geq ψ$ .

[F2]

Positive, negative, and truncated Sobolev functions: $ψ^{+} = \max {ψ, 0}$ is the class of the pointwise maximum, and on representatives $ψ^{+} \geq ψ$ pointwise.

[F3]

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences: a sequence converging in $L^{2} (Ω)$ has a subsequence converging almost everywhere on $Ω$ ; a sequence converging in $H_{0}^{1} (Ω)$ therefore has such a subsequence, since the $H_{0}^{1}$ norm dominates the $L^{2}$ norm (Integer-order Sobolev spaces and their norms).

[F4]

Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms: $H_{0}^{1} (Ω)$ is a closed linear subspace of $H^{1} (Ω)$ , and the almost-everywhere inequality $v \geq ψ$ is a condition on classes, independent of representatives.

[F5]

A norm-closed convex set is weakly sequentially closed: under the Axiom of Choice a convex norm closed subset of a normed space is weakly closed, hence weakly sequentially closed.

Proof

technique · direct

Given: The setting above, with $T ψ \leq 0$ and $K$ as defined.

1.1givenF1F2

(Nonemptiness) By [F1] the boundary condition gives $ψ^{+} \in H_{0}^{1} (Ω)$ , and by [F2] one has $ψ^{+} \geq ψ$ almost everywhere, so $ψ^{+} \in K$ .

1.2givenF4

(Convexity) Let $v, w \in K$ and $0 \leq t \leq 1$ . Then $t v + (1 - t) w \in H_{0}^{1} (Ω)$ because $H_{0}^{1} (Ω)$ is a linear subspace [F4], and $t v + (1 - t) w \geq ψ$ almost everywhere because this holds for $v$ and $w$ ; hence $t v + (1 - t) w \in K$ and $K$ is convex.

1.3givenF3F4

(Norm closedness) Let $(v_{j}) \subseteq K$ with $v_{j} \to v$ in $H_{0}^{1} (Ω)$ . By [F3] a subsequence converges to $v$ almost everywhere, and $v_{j} \geq ψ$ almost everywhere for every $j$ , so $v \geq ψ$ almost everywhere; since $H_{0}^{1} (Ω)$ is closed in $H^{1} (Ω)$ [F4], $v \in H_{0}^{1} (Ω)$ and hence $v \in K$ by [F4].

2.1step 1.1step 1.2step 1.3F5∎

(Weak sequential closedness and conclusion) By steps 1.1-1.3 the set $K$ is nonempty, convex and norm closed; [F5] therefore makes $K$ weakly closed, so in particular weakly sequentially closed: every weakly convergent sequence in $K$ has its limit in $K$ . This proves all the assertions; the choice principles enter only through the truncation interface of [F1] and the weak-closedness criterion [F5].

Depends on

Used by

Existence and uniqueness for the obstacle problem Theorem

Dependency tree · two levels

76 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

John Andersson, The Obstacle Problem, KTH lecture notes, 16 December 2015 (complete 52-page notes) (standard reference, not scraped)