CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An inward cusp blocks W^{1,3/2} extension

Statement refuted

Not every bounded connected open set is a Sobolev extension domain. In $R^{2}$ let $C = {(x, y) : x \geq 0, ∣ y ∣ \leq x^{2}}, Ω = B (0, 1) ∖ C .$ Then $Ω$ is a bounded connected open set with an inward cusp at the origin, and the branch $u = \arg (x + i y) / (2 π)$ , $0 < \arg < 2 π$ , of the argument on $Ω$ lies in $W^{1, 3 / 2} (Ω)$ but admits no extension to a class in $W^{1, 3 / 2} (R^{2})$ . Consequently $Ω$ is not a $W^{1, 3 / 2}$ -extension domain in the sense of Sobolev extension domains and extension operators. The obstruction is the exact $L^{3 / 2}$ summability threshold: the gradient of the argument has size $1 / (2 π r)$ , which is integrable to the power $3 / 2$ on $Ω$ but forces any extension to spend more than $c / x$ of vertical $L^{3 / 2}$ derivative energy on the gap of width $2 x^{2}$ at distance $x$ from the tip.

Facts & Assumptions

Given: the Axiom of Choice; the set $C = {(x, y) \in R^{2} : x \geq 0, ∣ y ∣ \leq x^{2}}$ ; the open set $Ω = B (0, 1) ∖ C$ ; the branch $u = \arg (x + i y) / (2 π)$ with $0 < \arg < 2 π$ ; and $p = 3 / 2$ .

[F1]

Under the assumed Axiom of Choice, ACL characterisation, $1 \leq p < \infty$ : $w \in W^{1, p} (Ω)$ if and only if $w \in L^{p} (Ω)$ and $w$ has a measurable ACL representative $w^{*}$ whose classical coordinate derivatives exist almost everywhere, are measurable, and lie in $L^{p} (Ω)$ ; in that case $\partial_{i} w^{*}$ represents $D_{i} w$ (The ACL characterisation of $W^{1, p}$ ).

[F2]

Polar coordinates: for Borel measurable $f \geq 0$ , $\int_{R^{2}} f d λ_{2} = \int_{0}^{\infty} \int_{S^{1}} f (r ω) r d σ (ω) d r$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F3]

Tonelli–Fubini for the identification of $λ_{2}$ with the completion of the product of the two Lebesgue measures (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

[F4]

Hölder's inequality on an interval of length $2 x^{2}$ : for $1 < r < \infty$ and $f \in L^{r}$ , $\int_{a}^{b} ∣ f ∣ \leq (b - a)^{1 / r^{'}} ∥ f ∥_{L^{r} (a, b)}$ with $r^{'}$ the conjugate exponent (Holder's inequality for integrals, including the endpoint cases).

[F5]

A $C^{1}$ function on an open set has its classical partial derivatives as weak derivatives there (Classical derivatives agree with weak derivatives), and the chain rule for the polar coordinate function $(x, y) \mapsto \arg (x + i y)$ gives $\nabla u = (- y, x) / (2 π (x^{2} + y^{2}))$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F6]

$W^{1, 3 / 2}$ -extension domain: $Ω$ is one exactly when there is a bounded linear operator $E : W^{1, 3 / 2} (Ω) \to W^{1, 3 / 2} (R^{2})$ with $(E u) ∣_{Ω} = u$ almost everywhere for every class $u$ (Sobolev extension domains and extension operators).

[F7]

For $1 \leq p < \infty$ , the norm is $∥ w ∥_{W^{1, p} (Ω)} = (∥ w ∥_{L^{p} (Ω)}^{p} + ∥ \partial_{x} w ∥_{L^{p} (Ω)}^{p} + ∥ \partial_{y} w ∥_{L^{p} (Ω)}^{p})^{1 / p}$ ; at $p = \infty$ it is $\max {∥ w ∥_{L^{\infty} (Ω)}, ∥ \partial_{x} w ∥_{L^{\infty} (Ω)}, ∥ \partial_{y} w ∥_{L^{\infty} (Ω)}}$ (Integer-order Sobolev spaces and their norms).

Choice use. The Axiom of Choice licenses the ACL interface [F1] and its Countable-Choice and Dependent-Choice prerequisites. Its countable instance also licenses the polar-coordinate and completed-product interfaces [F2]–[F3] and the Sobolev conventions. The vertical-section argument makes no further selections.

Counterexample

1.1F5given

On $Ω$ the branch $u$ is real-valued with $0 < u < 1$ , the function $u$ is $C^{\infty}$ , and by [F5] $∣ \nabla u (x, y) ∣ = \frac{1}{2 π \sqrt{x^{2} + y^{2}}} = \frac{1}{2 π r}$ at every point of $Ω$ ; $Ω$ is open and bounded. It is path connected: on every circle $∣ z ∣ = r < 1$ , the removed cusp occupies an arc around the positive real axis, while either complementary arc from a point of $Ω$ to $(- r, 0)$ stays in $Ω$ ; the negative real segment then joins $(- r, 0)$ to $(- 1 / 2, 0)$ .

2.1F2step 1.1

Integrability. Since $∣ u ∣ \leq 1$ and $Ω \subseteq B (0, 1)$ has finite area, $\int_{Ω} ∣ u ∣^{3 / 2} < \infty$ ; and [F2] gives $\int_{Ω} ∣ \nabla u ∣^{3 / 2} d x d y \leq (2 π)^{- 3 / 2} \int_{0}^{1} r^{- 3 / 2} 2 π r d r = (2 π)^{- 1 / 2} \int_{0}^{1} r^{- 1 / 2} d r = 2 (2 π)^{- 1 / 2} < \infty .$

3.1F1step 1.1step 2.1

Consequently $u \in W^{1, 3 / 2} (Ω)$ : the $C^{\infty}$ representative of step 1.1 is ACL with classical derivatives $\partial_{x} u, \partial_{y} u$ , which are measurable and, by step 2.1, lie in $L^{3 / 2} (Ω)$ , and $u \in L^{3 / 2} (Ω)$ ; the implication of [F1] applies.

4.1F1F3step 3.1

Suppose $U \in W^{1, 3 / 2} (R^{2})$ satisfies $U ∣_{Ω} = u$ almost everywhere, and take its ACL representative $U^{*}$ from [F1]. For almost every $x \in (0, 1 / 2)$ : the vertical section $U^{*} (x, \cdot)$ is absolutely continuous on the compact interval $[- \sqrt{1 - x^{2}}, \sqrt{1 - x^{2}}]$ ; since $U^{*} = u$ almost everywhere on $Ω$ while $u$ is continuous on each of the two open pieces of the section, $U^{*}$ agrees on each piece with the continuous function $y \mapsto u (x, y)$ , so the values at the two ends of the gap are $U^{*} (x, x^{2}) = \frac{\arctan x}{2 π}, U^{*} (x, - x^{2}) = 1 - \frac{\arctan x}{2 π} .$

5.1F4step 4.1

Gap energy. For those $x$ the difference of the two values of step 4.1 is $1 - \arctan (x) / π > 1 / 2$ , so by the fundamental theorem for the absolutely continuous section and [F4] $\frac{1}{2} < ∣ \int_{- x^{2}}^{x^{2}} \partial_{y} U^{*} (x, y) d y ∣ \leq (2 x^{2})^{1 / 3} (\int_{- x^{2}}^{x^{2}} ∣ \partial_{y} U^{*} ∣^{3 / 2} d y)^{2 / 3},$ hence $\int_{- x^{2}}^{x^{2}} ∣ \partial_{y} U^{*} (x, y) ∣^{3 / 2} d y \geq (1 / 2)^{3 / 2} (2 x^{2})^{- 1 / 2} = \frac{1}{4} x^{- 1} .$

6.1F1F3step 5.1

Integrating the lower bound of step 5.1 over $x \in (0, 1 / 2)$ gives $+ \infty$ , while Tonelli's theorem bounds the same double integral by $\int_{R^{2}} ∣ \partial_{y} U ∣^{3 / 2} < \infty$ , since $\partial_{y} U^{*}$ represents the $L^{3 / 2}$ class $\partial_{y} U$ by [F1]; this contradiction shows that no such $U$ exists.

7.1F6F7step 3.1step 6.1∎

Therefore $Ω$ is not a $W^{1, 3 / 2}$ -extension domain: if a bounded linear extension operator existed, [F6] applied to the class $u \in W^{1, 3 / 2} (Ω)$ of step 3.1 would produce exactly the extension $U$ excluded in step 6.1, with the norm bound of [F7] playing no role in the contradiction because the obstruction already lies in the membership.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

70 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

Juha Kinnunen, Sobolev Spaces (2026), Definition 3.42 and Theorem 1.25 (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (2020), Theorem 3.12 (standard reference, not scraped)