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The affine Dirichlet trace class is nonempty, convex and weakly closed

Statement

Assume the Axiom of Choice. Let $n \geq 2$ and let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain, $1 < p < \infty$ , and let $g \in W^{1 - 1 / p, p} (\partial Ω)$ lie in the trace range of $T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space, The $L^{p}$ trace operator on a bounded $C^{1}$ domain). Then the affine trace class $K_{g} : = {v \in W^{1, p} (Ω) : T v = g}$ is nonempty, convex, norm closed in $W^{1, p} (Ω)$ and weakly closed; moreover, for every right inverse $R$ of $T$ with $T R g = g$ , $K_{g} = R g + W_{0}^{1, p} (Ω)$ (A bounded right inverse of the trace, supported in a prescribed collar, The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure).

Facts & Assumptions

Given: The Axiom of Choice; $n \geq 2$ ; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $1 < p < \infty$ , and $g \in W^{1 - 1 / p, p} (\partial Ω)$ lying in the range of the trace operator $T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ of The $L^{p}$ trace operator on a bounded $C^{1}$ domain; the affine class $K_{g} : = {v \in W^{1, p} (Ω) : T v = g}$ .

[F1]

For the stated $n \geq 2$ and $1 < p < \infty$ , the trace operator $T : W^{1, p} (Ω) \to W^{θ, p} (\partial Ω)$ , $θ = 1 - 1 / p$ , is bounded and surjective onto the fractional Sobolev space (The sharp trace theorem: boundedness and range in the fractional space, The $L^{p}$ trace operator on a bounded $C^{1}$ domain).

[F2]

There is a bounded right inverse $R : W^{θ, p} (\partial Ω) \to W^{1, p} (Ω)$ with $T \circ R = i d$ ; it is not unique (A bounded right inverse of the trace, supported in a prescribed collar).

[F3]

$\ker T = W_{0}^{1, p} (Ω)$ , the $W^{1, p}$ -closure of $C_{c}^{\infty} (Ω)$ (The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure); in particular $W_{0}^{1, p} (Ω)$ is a linear subspace of $W^{1, p} (Ω)$ , which is a normed space for $∥ \cdot ∥_{W^{1, p} (Ω)}$ (Integer-order Sobolev spaces and their norms).

[F4]

Under the Axiom of Choice, every convex subset of a real or complex normed space that is closed in the norm topology is weakly closed (A norm-closed convex set is weakly sequentially closed).

Proof

technique · direct, by writing $K_g$ as a translate of the kernel of the trace and applying the closed-convex weak-closure lemma

1.1F1F2given

Nonemptiness. Since $g$ lies in the range of $T$ there is $u \in W^{1, p} (Ω)$ with $T u = g$ , so $K_{g} \neq \emptyset$ ; alternatively [F2] gives $R g \in K_{g}$ because $T R g = g$ .

1.2F2F3given

The class is a translate of the kernel. Fix a right inverse $R$ of $T$ , which exists by [F2] and satisfies $T R g = g$ . For $u \in W^{1, p} (Ω)$ one has $u \in K_{g} ⟺ T u = g ⟺ T (u - R g) = 0 ⟺ u - R g \in \ker T = W_{0}^{1, p} (Ω)$ , using linearity of $T$ and [F3]. Hence $K_{g} = R g + W_{0}^{1, p} (Ω)$ .

2.1F3step 1.2

Convexity. Let $u, v \in K_{g}$ and $λ \in [0, 1]$ . By step 1.2 the elements $u - R g$ and $v - R g$ lie in the linear subspace $W_{0}^{1, p} (Ω)$ , so $λ u + (1 - λ) v - R g = λ (u - R g) + (1 - λ) (v - R g) \in W_{0}^{1, p} (Ω)$ and therefore $λ u + (1 - λ) v \in K_{g}$ . Thus $K_{g}$ is convex.

2.2F1F3step 1.2

Norm closedness. The operator $T$ is bounded by [F1], hence continuous, and $K_{g} = T^{- 1} ({g})$ is the preimage of the singleton ${g}$ , which is closed in the normed space $W^{1 - 1 / p, p} (\partial Ω)$ ; a continuous preimage of a closed set is closed. So $K_{g}$ is closed in the norm topology of $W^{1, p} (Ω)$ .

3.1F4step 2.1step 2.2

Weak closedness. By steps 2.1 and 2.2 the set $K_{g}$ is convex and closed in the norm topology, so [F4] applies under the Axiom of Choice and $K_{g}$ is weakly closed.

4.1F2F3step 1.2step 3.1∎

The identity for an arbitrary right inverse. Let $R$ be any bounded right inverse of $T$ , so that $T R g = g$ . The argument of step 1.2 used only this identity and the kernel description [F3], so it gives $K_{g} = R g + W_{0}^{1, p} (Ω)$ for this $R$ as well. This, together with steps 1.1, 2.1 and 3.1, establishes every clause of the statement.

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Sources

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)
Francesco Paolo Maiale (course by Giovanni Alberti), Lecture Notes Calculus of Variations A, University of Pisa (last update 21 August 2019; complete 149-page notes) (standard reference, not scraped)