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Boundary point values are not a function of the interior $L^{p}$ class

Statement refuted

Assume the Axiom of Choice. The following two claims are false. (1) On a bounded $C^{1}$ domain $Ω$ , boundary values are a function of the interior class: for every $L^{p} (Ω)$ class the pointwise boundary values of a representative are determined by the class, so that classical restriction would descend to $L^{p} (Ω)$ . (2) Every $L^{p} (Ω)$ class is bounded near $\partial Ω$ , so that pointwise evaluation on $\partial Ω$ could be recovered from the interior class. In fact two functions with the same interior class can have different classical boundary restrictions, and a single $L^{2}$ class can be unbounded on every neighbourhood of the boundary; the Sobolev trace is defined on $W^{1, p}$ classes and does not assign a trace to every $L^{p}$ class.

Facts & Assumptions

Given: The Axiom of Choice; $n \geq 2$ ; the unit ball $Ω = B (0, 1)$ ; the functions $f = 0$ and $\tilde{f} = 1_{\partial Ω}$ on $\overline{Ω}$ ; the trace operator $T$ of The $L^{p}$ trace operator on a bounded $C^{1}$ domain; and $1 \leq p < \infty$ , $k \geq 0$ .

[F1]

$L^{p} (Ω)$ is the quotient of the $p$ -integrable measurable functions by almost-everywhere equality. Two such functions differing only on a null set define the same $L^{p}$ class; if that class belongs to $W^{k, p} (Ω)$ , they represent the same Sobolev element. The zero class belongs to every $W^{k, p} (Ω)$ , since all its weak derivatives are zero. (The space $L^{p} (μ)$ as the quotient by null functions, Integer-order Sobolev spaces and their norms)

[F2]

The unit sphere has zero ambient Lebesgue measure: the polar formula applied to its indicator has nonzero sections only at the radial singleton $r = 1$ , which has one-dimensional measure zero by the box formula. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

[F3]

For a $W^{1, p} (Ω)$ class admitting a representative continuous on $\overline{Ω}$ , the trace is that representative's classical restriction. (The trace agrees with classical restriction for continuous Sobolev functions)

[F4]

Assume the Axiom of Countable Choice. Polar coordinates give $\int_{B (0, 1)} h d x = \int_{0}^{1} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ for $h \geq 0$ Borel. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

Counterexample

On $Ω = B (0, 1)$ the functions $f = 0$ and $\tilde{f} = 1_{\partial Ω}$ differ only on the Lebesgue-null set $\partial Ω$ , so they represent the same element of $L^{p} (Ω)$ and of $W^{k, p} (Ω)$ for every $k, p$ ; yet their classical restrictions to $\partial Ω$ are the zero function and the constant-one function, which differ on the whole boundary, while $T f = T \tilde{f} = 0$ by The $L^{p}$ trace operator on a bounded $C^{1}$ domain.

1.1F1F2F3algebragiven

Two different boundary restrictions, one class. The set $\partial Ω$ has measure zero by [F2], so $f$ and $\tilde{f}$ agree off a null set; their restrictions to $Ω$ are both identically zero, so [F1] identifies their common $L^{p} (Ω)$ class with the zero element of $W^{k, p} (Ω)$ for every $k$ and $p$ . Their classical restrictions to $\partial Ω$ are $0$ and $1$ respectively, which differ at every point of $\partial Ω$ . Since $f$ is continuous on $\overline{Ω}$ , $T f = 0$ by [F3]; and $T \tilde{f} = T f$ because $\tilde{f}$ is a representative of the class of $f$ and $T$ is defined on classes. Hence the boundary values of a representative carry information invisible to $T$ .

1.2F4algebragiven

An $L^{2}$ class with no finite boundary values. On the unit ball $B (0, 1) \subset R^{n}$ put $g (x) : = (1 - ∣ x ∣)^{- 1 / 4}$ . Then $\int_{B} g^{2} d x = \int_{B} (1 - ∣ x ∣)^{- 1 / 2} d x = \int_{0}^{1} (1 - r)^{- 1 / 2} r^{n - 1} σ (S^{n - 1}) d r \leq σ (S^{n - 1}) \int_{0}^{1} (1 - r)^{- 1 / 2} d r = 2 σ (S^{n - 1}) < \infty$ by [F4], so $g \in L^{2} (B)$ . On the other hand $g (x) \to \infty$ as $∣ x ∣ \to 1$ , so $g$ is unbounded on every neighbourhood of $\partial B$ : no finite boundary values can be assigned from pointwise evaluation.

2.1step 1.1step 1.2algebragiven∎

Conclusion. Step 1.1 shows that two functions with the same interior class can have different classical boundary restrictions, while both have zero trace; step 1.2 shows that a single $L^{p}$ class need not be bounded near the boundary, so pointwise boundary evaluation is not a well-defined operation on $L^{p} (Ω)$ classes. For $W^{1, p}$ classes the Sobolev trace supplies boundary data independent of representatives; this does not extend pointwise evaluation to all $L^{p}$ classes.

Source notes

Laugesen's opening example (printed p. 62) is the function $(1 - ∣ x ∣)^{- 1 / 4}$ with infinite boundary values; Teschl's Problem set on traces (Problems 9.19 and 9.22, printed p. 211) records that classical restriction is not controlled by the interior $L^{p}$ norm, and Hunter's boundary-layer sequence (printed pp. 71-72) is the same failure in one dimension. The example above separates the two independent mechanisms: a null-set change of representative and an unbounded near-boundary profile.

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Sources

Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete 158-page graduate notes) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (archived 2025 author manuscript) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis, complete 242-page two-quarter notes) (standard reference, not scraped)

Boundary point values are not a function of the interior LpLp class