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Interior and Boundary Sobolev Elliptic Regularity — Examples

1 · Prerequisites

2 · Summary

These companions compute and stress-test the regularity theory of the main page. Poisson's equation with $L^{2}$ data gains exactly two interior derivatives for the constant-coefficient Laplacian, and bootstrapping a smooth datum on nested compact subsets produces a $C^{\infty}$ representative solving the equation pointwise. On the other side, the coefficient hypotheses are tested: a bounded discontinuous coefficient admits an $H^{1}$ weak solution with continuous flux that is not in $H^{2}$ , so bounded measurability cannot replace the Lipschitz hypothesis of the interior theorem, while a continuous piecewise-smooth (Lipschitz) coefficient gives an $H^{2}$ solution whose classical second derivative jumps at the interface, separating the weak $H^{2}$ scale from $C^{2}$ .

The boundary theory is bounded by counterexamples. A harmonic function on the slit disc is smooth inside and belongs to $H^{1}$ but not to $H^{2}$ , so interior regularity does not imply boundary regularity when the boundary fails to be a $C^{1}$ graph; on a reentrant sector the same mechanism produces the singular exponent $α = π / ω$ , with an explicit integer-order Sobolev threshold and a proved real-order threshold in the intrinsic Slobodeckij scale. At a convex corner of the square, smooth side data assigning different constants at a corner admit no solution continuous on the closure and no $H^{2}$ solution either, so compatibility of the data is not created by regularity. Finally, the $H^{2}$ estimate is shown to need its $L^{2}$ kernel term when the homogeneous problem has a nontrivial solution: on the unit disk, $L = - div ((1 / 4 - ∣ x ∣^{2} / 8) \nabla) - 1$ annihilates $u = 1 - ∣ x ∣^{2}$ , whose $H^{2}$ norm is positive while $∥ L u ∥_{L^{2}} = 0$ , and the interior cusp datum $∣ x ∣^{k + 1 / 2}$ shows that the two-derivative gain of the higher regularity theorem cannot be improved by smoothness of the coefficients alone.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Poisson's equation with $L^{2}$ data gains two interior derivatives

Example

Assume the Axiom of Choice where the existence of the weak solution is invoked; the regularity conclusion itself uses only Countable Choice. Let $Ω \subset R^{n}$ be a bounded open set, let $f \in L^{2} (Ω)$ , and let $u \in H_{0}^{1} (Ω)$ be a weak solution of the Dirichlet problem $- Δ u = f$ , that is, $a (u, v) = \int_{Ω} f \overline{v} d x$ for every $v \in H_{0}^{1} (Ω)$ with $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ (Weak Dirichlet solutions for a divergence-form operator). Assuming the Axiom of Choice, existence and uniqueness of such a $u$ are supplied by Existence and uniqueness for the weak Dirichlet Poisson problem when $Ω$ is nonempty; if $Ω = \emptyset$ , the zero class is the unique weak solution directly. The verification below uses only that $u$ is a weak solution. Then $u \in H_{l o c}^{2} (Ω)$ , and for every open $Ω^{'} ⋐ Ω$ there is $C = C (n, Ω^{'}, Ω)$ with $∥ u ∥_{H^{2} (Ω^{'})} \leq C (∥ f ∥_{L^{2} (Ω)} + ∥ u ∥_{L^{2} (Ω)}) :$ $L^{2}$ data gain two interior derivatives for the constant-coefficient Laplacian, and no boundary regularity of $Ω$ enters the interior conclusion.

Facts & Assumptions

Given: The Axiom of Choice (for the existence statement only); a bounded open set $Ω \subset R^{n}$ ; $f \in L^{2} (Ω)$ ; and a weak solution $u \in H_{0}^{1} (Ω)$ of $- Δ u = f$ in the sense of Weak Dirichlet solutions for a divergence-form operator.

[F1]

For $f \in L^{2} (Ω)$ with $Ω$ bounded, the weak Dirichlet formulation reads $a (u, v) = \int_{Ω} f \overline{v} d x$ for every $v \in H_{0}^{1} (Ω)$ ; every such $u$ is a local weak solution of $- Δ u = f$ on $Ω$ , because $C_{c}^{\infty} (Ω) \subseteq H_{0}^{1} (Ω)$ and the local definition tests the smaller class. (Weak Dirichlet solutions for a divergence-form operator, Local weak solutions of a divergence-form operator)

[F2]

The Laplacian $L = - Δ$ is the divergence-form operator with $a^{i j} = δ^{i j}$ , $b^{i} = 0$ , $c = 0$ : the coefficients are constant, hence in $W^{1, \infty} (Ω)$ with $∥ D a^{i j} ∥_{\infty} = 0$ and $M_{1} = 0$ , uniformly elliptic with $θ = 1$ , and $M_{a} = 1$ , $M_{b} = M_{c} = 0$ . (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F3]

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open and let $L, a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with $a^{i j} \in W^{1, \infty} (Ω)$ , $∥ D a^{i j} ∥_{\infty} \leq M_{1}$ , and $f \in L_{l o c}^{2} (Ω)$ . If $u \in H^{1} (Ω)$ is a local weak solution of $L u = f$ on $Ω$ , then $u \in H_{l o c}^{2} (Ω)$ and for every pair of open sets $Ω^{'} ⋐ Ω^{''} ⋐ Ω$ the theorem gives the interior estimate of Interior $H^{2}$ regularity for divergence-form equations. When $f \in L^{2} (Ω)$ , the estimate on $Ω^{'}$ is bounded by the global-norm estimate displayed in the statement because $Ω^{''} \subseteq Ω$ ; its constant may be written $C (n, θ, M_{a}, M_{b}, M_{c}, M_{1}, Ω^{'}, Ω)$ after fixing such an intermediate $Ω^{''}$ from $Ω^{'}$ and $Ω$ .

[F4]

Assume the Axiom of Choice. If $Ω$ is nonempty, every $F \in H^{- 1} (Ω)$ has exactly one weak solution of the Dirichlet problem $- Δ u = F$ with zero boundary values; for $F (v) = \int_{Ω} f \overline{v} d x$ with $f \in L^{2} (Ω)$ this supplies existence and uniqueness in the example. If $Ω = \emptyset$ , then $H^{- 1} (Ω) = {0}$ and the zero class is the unique weak solution directly. (Existence and uniqueness for the weak Dirichlet Poisson problem)

Verification

1.1F1F2given

Hypothesis check for [F3]. By [F2] the Laplacian has constant coefficients and $∥ D a^{i j} ∥_{\infty} = 0$ , so it is admissible with $M_{1} = 0$ , $θ = 1$ , $M_{a} = 1$ and $M_{b} = M_{c} = 0$ ; since $Ω$ is bounded and $f \in L^{2} (Ω)$ , also $f \in L_{l o c}^{2} (Ω)$ ; and by [F1] the weak solution $u$ is a local weak solution of $- Δ u = f$ on $Ω$ .

2.1F3F4step 1.1algebra∎

Fix any open $Ω^{'} ⋐ Ω$ and choose an open $Ω^{''}$ with $Ω^{'} ⋐ Ω^{''} ⋐ Ω$ . By step 1.1 the solution satisfies the hypotheses of [F3], so the theorem gives $u \in H^{2} (Ω^{'})$ and $∥ u ∥_{H^{2} (Ω^{'})} \leq C (n, θ, M_{a}, M_{b}, M_{c}, M_{1}, Ω^{'}, Ω^{''}) (∥ f ∥_{L^{2} (Ω^{''})} + ∥ u ∥_{L^{2} (Ω^{''})}) .$ Since $Ω^{''} \subseteq Ω$ , both local norms are bounded by the global norms in the statement. Fixing the intermediate set as a function of $Ω^{'}$ and $Ω$ therefore gives $C = C (n, θ, M_{a}, M_{b}, M_{c}, M_{1}, Ω^{'}, Ω)$ ; for the Laplacian all coefficient parameters are absolute and $M_{1} = 0$ , so $C = C (n, Ω^{'}, Ω)$ . Boundary regularity of $Ω$ is not part of the hypotheses of [F3], so it is not used; existence of $u$ is the only place the Axiom of Choice enters, through [F4] (with the empty-domain case handled directly). This proves the claimed two-derivative interior gain.

Source notes

Hunter's motivating computation for the Laplacian and Theorem 4.27 (printed pp. 110-114, read in full) state the interior estimate with $∥ f ∥_{L^{2} (Ω)}$ and no boundary hypothesis; Laugesen's Theorem 5.6 (printed p. 108) is the same interior $H^{2}$ statement. The example isolates the constant-coefficient case: the coefficient constants in the estimate are absolute, so the constant depends only on $n, Ω^{'}, Ω$ , and the estimate does not improve when $Ω$ is smoother.

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A piecewise-smooth coefficient gives an $H^{2}$ solution that is not twice classically differentiable

Example

Assume Countable Choice. On $Ω = (- 1, 1)$ let $a (x) = {\begin{cases} 2 + x, & x \leq 0, \\ 2 + 2 x, & x > 0, \end{cases}$ a continuous, piecewise smooth coefficient with $1 \leq a \leq 4$ and $a \in W^{1, \infty} (- 1, 1)$ whose derivative jumps from $1$ to $2$ at $0$ , and define $u (x) = \int_{0}^{x} \frac{d t}{a (t)}$ , the absolutely continuous primitive with $u (0) = 0$ . Then $u \in H^{2} (- 1, 1)$ , the flux is $a u^{'} \equiv 1$ , and $u$ is a local weak solution of $- (a u^{'})^{'} = 0$ in the sense of Local weak solutions of a divergence-form operator. Explicitly $u^{''} (x) = - \frac{a^{'} (x)}{a (x)^{2}} = {\begin{cases} - \frac{1}{(2 + x)^{2}}, & x < 0, \\ - \frac{2}{(2 + 2 x)^{2}}, & x > 0, \end{cases}$ with one-sided limits $- 1 / 4$ at $0^{-}$ and $- 1 / 2$ at $0^{+}$ ; consequently the one-sided difference quotients of $u^{'}$ at $0$ have the distinct limits $- 1 / 4$ and $- 1 / 2$ , the second classical derivative of $u$ at $0$ does not exist, and $u \notin C^{2} (- 1, 1)$ . Thus the weak $H^{2}$ conclusion of Interior $H^{2}$ regularity for divergence-form equations holds for this Lipschitz coefficient while classical twice differentiability fails: weak $H^{2}$ regularity is genuinely weaker than $C^{2}$ .

Facts & Assumptions

Given: The coefficient $a$ above and the primitive $u (x) = \int_{0}^{x} d t / a (t)$ ; the operator $L u = - (a u^{'})^{'}$ with $a^{11} = a$ , $b = c = 0$ .

[F1]

$u$ is locally absolutely continuous, $u (0) = 0$ , and its a.e. derivative is the integrand: $u^{'} (x) = 1 / a (x)$ for a.e. $x \in (- 1, 1)$ ; more precisely $u (x) = \ln (2 + x) - \ln 2$ for $x \leq 0$ and $u (x) = \frac{1}{2} \ln (1 + x)$ for $x \geq 0$ , and these formulas are differentiable with the stated values of $1 / a$ on each side, agreeing at $0$ with value $1 / 2$ . (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ )

[F2]

A class $u \in H^{1} (Ω)$ is a local weak solution of $L u = f$ on $Ω$ if $a (u, v) = \int_{Ω} f \overline{v}$ for every $v \in C_{c}^{\infty} (Ω)$ ; for $L = - (a u^{'})^{'}$ and $f = 0$ this is $\int_{Ω} a u^{'} \overline{v^{'}} d x = 0$ for every $v \in C_{c}^{\infty} (Ω)$ . (Local weak solutions of a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F3]

The operator $- (a u^{'})^{'}$ is uniformly elliptic with $a^{11} = a$ : $1 \leq a \leq 4$ gives $θ = 1$ and $M_{a} = 4$ , and $b = c = 0$ ; $a$ is continuous and piecewise $C^{1}$ with bounded derivative, hence $a \in W^{1, \infty} (- 1, 1)$ . (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F4]

For a continuous function $w$ that is $C^{1}$ on each of $(- 1, 0)$ and $(0, 1)$ with bounded one-sided derivatives, integrate $w φ^{'}$ on the two half-intervals. The terms at $0$ cancel because its one-sided values agree; the outer terms vanish for $φ \in C_{c}^{\infty} (- 1, 1)$ . Thus its piecewise derivative is its weak derivative. This argument applies to the coefficient $a$ and to $w = 1 / a$ here. Almost-everywhere differentiability and integrability of the classical derivative alone would not suffice for a general function. (Weak derivative of a locally integrable function)

Verification

1.1F1algebragiven

The explicit primitive. By [F1], for $x < 0$ one has $u (x) = \int_{0}^{x} d t / (2 + t) = \ln (2 + x) - \ln 2$ , and for $x > 0$ one has $u (x) = \int_{0}^{x} d t / (2 + 2 t) = \frac{1}{2} \ln (1 + x)$ ; both formulas give $u (0) = 0$ and both one-sided derivatives at $0$ equal $1 / 2$ , so $u$ is differentiable at $0$ and $u^{'} = 1 / a$ on $(- 1, 1)$ . In particular $u^{'}$ is continuous and $u \in H^{1} (- 1, 1) \cap L^{\infty} (- 1, 1)$ .

1.2F1F4algebra

The second derivative. On $(- 1, 0)$ and on $(0, 1)$ the coefficient is smooth and $u^{''} = - a^{'} / a^{2}$ , namely $- 1 / (2 + x)^{2}$ and $- 2 / (2 + 2 x)^{2}$ respectively; both expressions are bounded in absolute value by $1$ , so $u^{''} \in L^{\infty} (- 1, 1)$ . Moreover $∣ 1 / a (x) - 1 / a (y) ∣ \leq ∣ a (x) - a (y) ∣ \leq 2 ∣ x - y ∣$ because $a \geq 1$ and $a$ is Lipschitz with constant $2$ ; hence $u^{'} = 1 / a$ is Lipschitz and continuous at $0$ . Applying the piecewise test calculation of [F4] shows directly that the displayed bounded piecewise derivative is its weak derivative. Consequently $u \in W^{2, \infty} (- 1, 1) \subseteq H^{2} (- 1, 1)$ , and the displayed formula for $u^{''}$ is the weak second derivative.

2.1F2F3step 1.1algebragiven

The weak equation. Since $a u^{'} \equiv 1$ by step 1.1, for every $v \in C_{c}^{\infty} (- 1, 1)$ one has $\int_{- 1}^{1} a u^{'} \overline{v^{'}} d x = \int_{- 1}^{1} \overline{v^{'}} d x = 0$ , the last integral vanishing because $v$ is compactly supported in $(- 1, 1)$ . Hence the identity of [F2] holds with $f = 0$ ; its hypothesis $u \in H^{1}$ is met by step 1.1. Thus $u$ is the local weak solution of $- (a u^{'})^{'} = 0$ , and by [F3] the operator has $θ = 1$ , $M_{a} = 4$ .

3.1F1step 1.2algebra∎

Failure of classical twice differentiability. By step 1.2 the one-sided limits of $u^{''}$ at $0$ are $- 1 / 4$ at $0^{-}$ and $- 1 / 2$ at $0^{+}$ ; equivalently, the difference quotients of $u^{'}$ have these one-sided limits, since for $h < 0$ one has $(u^{'} (h) - u^{'} (0)) / h = - 1 / (2 (2 + h)) \to - 1 / 4$ and for $h > 0$ one has $(u^{'} (h) - u^{'} (0)) / h = - 1 / (2 + 2 h) \to - 1 / 2$ . Hence $u^{'}$ is not differentiable at $0$ and $u \notin C^{2} (- 1, 1)$ , while $u \in H^{2} (- 1, 1)$ by step 1.2.

Source notes

This is the Lipschitz-coefficient threshold case of the interior $H^{2}$ theorem of Interior $H^{2}$ regularity for divergence-form equations: Teschl's Lemma 10.16 (printed p. 240) assumes exactly $A \in W^{1, \infty}$ , Hunter's Theorem 4.27 (printed p. 112) assumes $C^{1}$ coefficients, and both give $H_{l o c}^{2}$ while saying nothing about $C^{2}$ . The computation above is deliberately self-contained: it verifies $u \in W^{2, \infty} \subseteq H^{2}$ directly from the explicit primitive, so the failure of $C^{2}$ at the derivative jump of $a$ is separated from the regularity theorem rather than resting on it.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Interior regularity does not imply boundary regularity

Statement refuted

Assume Countable Choice. Interior smoothness of a weak solution automatically forces $H^{2}$ up to the boundary of its domain, so that the interior tangential and normal estimates suffice at every boundary point.

Facts & Assumptions

Given: The slit plane $S = C ∖ {x \in R : x \leq 0}$ , the slit disc $Ω = S \cap B_{1} (0) \subset R^{2}$ , the principal square-root biholomorphism $R_{2} (z) = \exp (Log z / 2)$ of the published slit plane theorem, and the function $u (z) = Im R_{2} (z)$ .

[F1]

$R_{2}$ is a biholomorphism from $S$ onto the sector $V_{2} = {r e^{i θ} : r > 0, - π / 2 < θ < π / 2}$ , with inverse $w \mapsto w^{2}$ ; in polar coordinates $z = r e^{i θ}$ with $θ \in (- π, π)$ one has $u (z) = r^{1 / 2} \sin (θ / 2)$ , and $u$ is $C^{\infty}$ and harmonic on $S$ . (A slit-plane root branch biholomorphically parametrizes a sector, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair)

[F2]

A class $u \in H^{1} (Ω)$ is a local weak solution of $- Δ u = 0$ on $Ω$ if $\int_{Ω} \nabla u \cdot \overline{\nabla v} d x = 0$ for every $v \in C_{c}^{\infty} (Ω)$ , equivalently for every $v \in H_{0}^{1} (Ω_{2})$ with $Ω_{2} ⋐ Ω$ bounded. (Local weak solutions of a divergence-form operator)

[F3]

Assume Countable Choice. If $a \in W^{1, 2} (U)$ and $b \in W^{1, 2} (U)$ for an open set $U \subseteq R^{n}$ , and at least one of them is compactly supported in $U$ , then $\int_{U} a D_{i} b d x = - \int_{U} b D_{i} a d x$ for every coordinate $i$ , the integrals being bilinear (no conjugation) and absolutely convergent. (Integration by parts for dual-exponent Sobolev functions)

[F4]

For $u = r^{1 / 2} \sin (θ / 2)$ one has $u_{r} = \frac{1}{2} r^{- 1 / 2} \sin (θ / 2)$ and $u_{θ} = \frac{1}{2} r^{1 / 2} \cos (θ / 2)$ , so $∣ \nabla u ∣^{2} = u_{r}^{2} + r^{- 2} u_{θ}^{2} = \frac{1}{4} r^{- 1}$ and $u_{r r} = - \frac{1}{4} r^{- 3 / 2} \sin (θ / 2)$ ; the polar-coordinate formula gives $\int_{Ω} f d x = \int_{0}^{1} \int_{- π}^{π} f (r \cos θ, r \sin θ) r d θ d r$ for every integrable $f$ , the slit lying in the boundary and being ${r > 0, θ = \pm π}$ , a polar-coordinate null set. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F5]

The boundary of $Ω$ is not locally the graph of a $C^{1}$ function at $0$ : every neighbourhood of $0$ meets both components of $Ω \cap {x_{2} \neq 0}$ and the slit ${x_{2} = 0, x_{1} < 0}$ lies in $\partial Ω$ , so the defining graph condition of a bounded $C^{k}$ domain fails at $0$ (and along the slit). (Bounded C^k domains and boundary charts)

Counterexample

1.1F1algebragiven

The function is smooth and harmonic inside the domain. By [F1], $R_{2}$ is holomorphic on the slit plane $S \supseteq Ω$ and $u = Im R_{2}$ has polar form $r^{1 / 2} \sin (θ / 2)$ with $θ \in (- π, π)$ ; since a holomorphic function is $C^{\infty}$ in its complex variable, $u \in C^{\infty} (Ω)$ and, by the published component theorem, $u$ is harmonic on $Ω$ : $Δ u = 0$ pointwise.

1.2F4algebra

The function lies in $H^{1}$ . By [F4], $∣ u ∣ \leq r^{1 / 2}$ and $∣ \nabla u ∣^{2} = \frac{1}{4} r^{- 1}$ on $Ω$ , so $\int_{Ω} (∣ u ∣^{2} + ∣ \nabla u ∣^{2}) d x \leq \int_{0}^{1} (r + \frac{1}{4} r^{- 1}) 2 π r d r = 2 π \int_{0}^{1} (r^{2} + \frac{1}{4}) d r$ , which is finite; hence $u \in H^{1} (Ω)$ .

2.1F2F3step 1.1algebra

The function is a local weak solution. Fix $v \in C_{c}^{\infty} (Ω)$ , write $v = v_{1} + i v_{2}$ with real-valued $v_{1}, v_{2}$ , put $d = dist (supp v, \partial Ω) > 0$ and $U = {x : dist (x, supp v) < d / 2}$ , an open bounded set with $\overline{U} \subseteq Ω$ and $supp v \subset U$ . On $U$ the real function $u$ is $C^{\infty}$ by step 1.1, so each real class $\partial_{i} u$ lies in $W^{1, 2} (U)$ , while $v_{1}, v_{2} \in C_{c}^{\infty} (U)$ . Applying [F3] on $U$ with $a = \partial_{i} u$ and $b = v_{j}$ gives $\int_{U} \nabla u \cdot \nabla v_{j} = - \int_{U} v_{j} Δ u = 0$ for each $j = 1, 2$ . Since $\overline{v} = v_{1} - i v_{2}$ , the original pairing is $\int_{U} \nabla u \cdot \nabla v_{1} - i \int_{U} \nabla u \cdot \nabla v_{2} = 0$ . By [F2], $u$ is a local weak solution of $- Δ u = 0$ on $Ω$ .

2.2F4step 1.2algebra

Membership in $H^{2}$ fails at the slit tip. By [F4], For the full Cartesian Hessian $H u = (\partial_{i} \partial_{j} u)_{i, j}$ , $∣ H u ∣ \geq ∣ u_{r r} ∣ = \frac{1}{4} r^{- 3 / 2} ∣ \sin (θ / 2) ∣$ , so $\int_{Ω} ∣ H u ∣^{2} d x \geq \frac{1}{16} \int_{0}^{1} \int_{- π}^{π} r^{- 3} \sin^{2} (θ / 2) r d θ d r = \frac{π}{16} \int_{0}^{1} r^{- 2} d r = + \infty$ ; hence $u \notin H^{2} (Ω)$ and no representative of $u$ is $H^{2}$ up to the boundary point $0$ .

3.1F5step 1.1step 2.2algebra∎

The failure is a boundary phenomenon, not an interior one. Every compactly contained open $Ω^{'} ⋐ Ω$ has positive distance from $0$ and from the slit, and there $u$ is $C^{\infty}$ and harmonic by step 1.1, so $u \in H_{l o c}^{2} (Ω)$ : the interior theorem applies on each $Ω^{'}$ and is not contradicted. The boundary $\partial Ω$ fails the graph condition at $0$ by [F5], so the global boundary hypotheses are unavailable exactly where the $H^{2}$ integral of step 2.2 diverges. Thus interior smoothness does not imply boundary regularity.

Source notes

Teschl's Example 10.1 and the surrounding discussion (printed p. 242) exhibit reentrant boundary points at which the harmonic model function $r^{α} \sin (α θ)$ is in $H^{1}$ but not $H^{2}$ ; the slit disc used here is the limiting case of interior angle $2 π$ with $α = 1 / 2$ , and the square-root biholomorphism of the published slit-plane theorem supplies the harmonicity without any polar-coordinate Laplacian computation. Laugesen's Theorem 5.10 (printed pp. 112-113) states the global boundary estimate under a $C^{2}$ boundary hypothesis, which is exactly what fails here at the slit. The scaffold's earlier witness on the punctured disc was replaced: the function $\log ∣ x ∣$ there is not in $H^{1}$ , so it is not an admissible weak solution and cannot witness the failure; the slit geometry is the minimal correct witness with the same role on this page.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Boundary $H^{2}$ regularity needs domain regularity

Statement refuted

Assume Countable Choice. In the global $H^{2}$ Dirichlet theorem the bounded $C^{2}$ boundary hypothesis can be replaced by mere Lipschitz regularity: on every bounded Lipschitz domain in $R^{2}$ , every zero-boundary weak solution of $- Δ u = f$ with $f \in L^{2}$ lies in $H^{2} (Ω)$ .

The reentrant-sector factor $v = r^{π / ω} \sin (π θ / ω)$ is in $H^{1} (S_{ω}) ∖ H^{2} (S_{ω})$ and is locally weakly harmonic. Multiplying it by a smooth cutoff equal to $1$ near the vertex and $0$ near the circular boundary gives a zero-boundary weak solution with $L^{2}$ forcing that is still not in $H^{2}$ .

Facts & Assumptions

Given: A number $ω \in (π, 2 π)$ , the reentrant sector $S_{ω} = {(r \cos θ, r \sin θ) : 0 < r < 1, 0 < θ < ω}$ , the exponent $α = π / ω \in (0, 1)$ , the singular harmonic function $v (r, θ) = r^{α} \sin (α θ)$ , and a smooth cutoff $χ$ on $R^{2}$ that equals $1$ on $B_{1 / 2} (0)$ and is supported in $B_{3 / 4} (0)$ . Put $u = χ v$ on $S_{ω}$ .

[F1]

For $g \in L_{l o c}^{2} (S_{ω})$ , a class $w \in H^{1} (S_{ω})$ is a local weak solution of $- Δ w = g$ if $\int_{S_{ω}} \nabla w \cdot \overline{\nabla φ} d x = \int_{S_{ω}} g \overline{φ} d x$ for every $φ \in C_{c}^{\infty} (S_{ω})$ ; if also $g \in L^{2} (S_{ω})$ and $w \in H_{0}^{1} (S_{ω})$ , density extends this identity to every $H_{0}^{1}$ test and gives the zero-boundary weak Dirichlet solution. (Local weak solutions of a divergence-form operator, Weak Dirichlet solutions for a divergence-form operator, Zero-boundary Sobolev space as a norm closure)

[F2]

Assume Countable Choice. If $a, b \in W^{1, 2} (U)$ on an open set $U \subseteq R^{n}$ and at least one of them is compactly supported in $U$ , then $\int_{U} a D_{i} b d x = - \int_{U} b D_{i} a d x$ for every coordinate $i$ , bilinearly and absolutely convergently. (Integration by parts for dual-exponent Sobolev functions)

[F3]

For every $C^{2}$ function $w$ on an open subset of the punctured plane, the chain rule applied to $x = r \cos θ$ , $y = r \sin θ$ gives $w_{x x} + w_{y y} = w_{r r} + \frac{1}{r} w_{r} + \frac{1}{r^{2}} w_{θ θ},$ because $r_{x}^{2} + r_{y}^{2} = 1$ , $θ_{x}^{2} + θ_{y}^{2} = r^{- 2}$ , $r_{x} θ_{x} + r_{y} θ_{y} = 0$ , $r_{x x} + r_{y y} = r^{- 1}$ and $θ_{x x} + θ_{y y} = 0$ where $r > 0$ . (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F4]

The polar-coordinate formula $\int_{S_{ω}} f d x = \int_{0}^{ω} \int_{0}^{1} f (r \cos θ, r \sin θ) r d r d θ$ holds for every integrable $f$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F5]

$S_{ω}$ is a bounded Lipschitz domain: after rotating the exterior angle bisector to the upward vertical direction, its boundary near the vertex is the graph of the Lipschitz function $t \mapsto \cot ((2 π - ω) / 2) ∣ t ∣$ and $S_{ω}$ is the region below that graph. At the vertex the boundary is not the graph of any $C^{1}$ function: the two radial edges meet there at interior angle $ω > π$ , so the defining chart condition of a bounded $C^{1}$ (hence $C^{1, 1}$ or $C^{2}$ ) domain fails. The circular arc is smooth, and at its two intersections with the radial edges the pieces meet transversely, giving ordinary Lipschitz corner charts. (Bounded C^k domains and boundary charts)

[F6]

The coefficients of $- Δ$ are $a^{i j} = δ^{i j}$ , $b = c = 0$ and the ellipticity constant is $θ = 1$ . (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F7]

Smooth cutoffs exist for $χ$ , the radial truncations $ρ_{ϵ}$ and the angular truncations $η_{δ}$ : the ball bump is used for $χ$ , and the compact-set bump supplies the one-dimensional cutoffs on radial and angular intervals. (A smooth bump between concentric Euclidean balls, A Euclidean bump for a compact set inside an open set)

[F8]

On a bounded $C^{2}$ domain in dimension $n \geq 2$ , every weak solution $u \in H_{0}^{1} (Ω)$ of $- Δ u = f$ with $f \in L^{2} (Ω)$ lies in $H^{2} (Ω)$ with $∥ u ∥_{H^{2}} \leq C (∥ f ∥_{L^{2}} + ∥ u ∥_{L^{2}})$ . (Global $H^{2}$ Dirichlet regularity)

Counterexample

1.1F3algebragiven

The singular factor is harmonic. For $v = r^{α} \sin (α θ)$ one has $v_{r r} = α (α - 1) r^{α - 2} \sin (α θ)$ , $v_{r} = α r^{α - 1} \sin (α θ)$ and $v_{θ θ} = - α^{2} r^{α} \sin (α θ)$ ; substituting into the polar formula of [F3] gives $Δ v = (α (α - 1) + α - α^{2}) r^{α - 2} \sin (α θ) = 0$ . It vanishes on both radial edges.

1.2F4F7algebra

The cutoff solution lies in $H_{0}^{1}$ . The cutoff $u = χ v$ has the same $H^{1}$ singularity near $0$ , is zero near $r = 1$ , and vanishes on the two radial edges. For $ϵ > 0$ choose a smooth radial cutoff $ρ_{ϵ}$ that is zero for $r \leq ϵ$ , one for $r \geq 2 ϵ$ , and satisfies $∣ D ρ_{ϵ} ∣ \leq C / ϵ$ . For $δ > 0$ choose a smooth angular cutoff $η_{δ}$ that vanishes within angular distance $δ$ of the two radial edges, equals one beyond distance $2 δ$ , and satisfies $∣ D η_{δ} ∣ \leq C / (r δ)$ in its transition strips. Then $u_{ϵ, δ} = ρ_{ϵ} η_{δ} u$ lies in $C_{c}^{\infty} (S_{ω})$ . Near the vertex $∣ u ∣ \leq C r^{α}$ and $∣ D u ∣ \leq C r^{α - 1}$ , so polar integration bounds the squared $H^{1}$ error from $ρ_{ϵ}$ by $C ϵ^{2 α}$ . For fixed $ϵ$ , the error from $η_{δ}$ tends to zero as $δ ↓ 0$ : near each edge $∣ u ∣ \leq C r^{α} d (θ)$ and $∣ D u ∣ \leq C r^{α - 1}$ , and the derivative-cutoff term has squared integral at most $C_{ϵ} δ$ . Choose $δ = δ (ϵ)$ so this second error tends to zero as $ϵ ↓ 0$ . Thus $u_{ϵ, δ (ϵ)} \to u$ in $H^{1}$ , proving $u \in H_{0}^{1} (S_{ω})$ .

2.1F4step 1.1algebra

The singular factor lies in $H^{1}$ but not $H^{2}$ . Its polar derivatives give $∣ \nabla v ∣^{2} = α^{2} r^{2 α - 2}$ and $∣ v ∣ \leq r^{α}$ , so by [F4] $\int_{S_{ω}} (∣ v ∣^{2} + ∣ \nabla v ∣^{2}) d x \leq ω (\frac{1}{2 α + 2} + \frac{α^{2}}{2 α}) < \infty .$ Thus $v \in H^{1} (S_{ω})$ . Its radial second derivative has squared integral $\int_{S_{ω}} ∣ v_{r r} ∣^{2} d x = α^{2} (1 - α)^{2} (\int_{0}^{ω} \sin^{2} (α θ) d θ) (\int_{0}^{1} r^{2 α - 3} d r) = + \infty,$ since $0 < α < 1$ . If all Cartesian second derivatives were in $L^{2}$ , then $v_{r r} = D^{2} v [e_{r}, e_{r}]$ would be in $L^{2}$ as well (the radial direction $e_{r}$ is a unit vector), a contradiction. Hence $v \notin H^{2} (S_{ω})$ .

2.2F3F7step 1.1algebra

The forcing is square-integrable. The function $u = χ v$ is smooth in the sector, and $f : = - Δ u$ vanishes wherever $χ$ is constant because $Δ v = 0$ . The derivatives of $χ$ are supported in the annulus $1 / 2 \leq r \leq 3 / 4$ , where $v$ and its derivatives are bounded. Therefore $f \in L^{2} (S_{ω})$ .

3.1F1step 1.1step 2.1algebra

The uncut factor is locally weakly harmonic. For any $φ \in C_{c}^{\infty} (S_{ω})$ , its support lies in a compact subset of the open sector where $v$ is smooth. Integration by parts there and $Δ v = 0$ give $\int_{S_{ω}} \nabla v \cdot \overline{\nabla φ} d x = 0$ . Thus $v$ is the local weak solution recorded in the statement.

3.2F1F2F5F6step 1.2step 2.2algebra

The weak equation and boundary condition. For every $φ \in C_{c}^{\infty} (S_{ω})$ , integration by parts on a neighborhood of its compact support gives $\int_{S_{ω}} \nabla u \cdot \overline{\nabla φ} d x = \int_{S_{ω}} f \overline{φ} d x$ . By step 1.2, $u \in H_{0}^{1} (S_{ω})$ ; both sides are continuous in the $H_{0}^{1}$ norm because $f \in L^{2}$ and the principal form is bounded. Density extends the identity to every $H_{0}^{1}$ test. Thus $u$ is a zero-boundary weak Dirichlet solution of $- Δ u = f$ .

3.3F4step 2.1step 2.2algebra

Failure of $H^{2}$ . On $B_{1 / 2} (0) \cap S_{ω}$ , $u = v$ , so the divergent radial second-derivative integral of step 2.1 also occurs for $u$ . As there $u_{r r} = D^{2} u [e_{r}, e_{r}]$ , this precludes $u \in H^{2} (S_{ω})$ .

4.1F5F8step 1.2step 2.2step 3.2step 3.3algebra∎

Lipschitz is not enough. By [F5], $S_{ω}$ is bounded Lipschitz but not $C^{1}$ at its vertex. Steps 1.2 and 2.2--3.2 give a zero-boundary weak solution with $f \in L^{2}$ , while step 3.3 shows that it is not in $H^{2}$ . The $C^{2}$ hypothesis of [F8] therefore cannot be replaced by Lipschitz regularity, even for the Laplacian, smooth forcing and zero boundary data.

Source notes

This is [T] Example 10.1 (printed p. 242) with the sector angle $ω > π$ and the singular exponent $α = π / ω$ ; Teschl uses it to show $u \notin H^{2} (S_{ω})$ . Laugesen's Theorem 5.10 (printed p. 112) is the global estimate under a $C^{2}$ boundary hypothesis. The scaffold's statements of the local weak solution and of the IBP lemma are realised here by the C_c^\infty definition and the published Sobolev integration-by-parts lemma, so no boundary-smoothness theorem is used in verifying the weak equation.

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Bounded discontinuous elliptic coefficients need not give $H^{2}$ solutions

Statement refuted

Assume Countable Choice. Bounded measurable uniformly elliptic coefficients together with $f = 0$ force every $H^{1}$ weak solution of $- (a^{i j} D_{j} u)_{i}^{'} = f$ into $H_{l o c}^{2}$ , without any regularity hypothesis on the coefficients.

Facts & Assumptions

Given: The interval $Ω = (- 1, 1)$ ; the coefficient $a (x) = {\begin{cases} 1, & x < 0, \\ 2, & x > 0, \end{cases}$ the primitive-shaped function $u (x) = {\begin{cases} x, & x < 0, \\ x / 2, & x > 0, \end{cases}$ and the operator $L u = - (a u^{'})^{'}$ , so that $a^{11} = a$ and $b = c = 0$ .

[F1]

A class $u \in H^{1} (- 1, 1)$ is a local weak solution of $L u = 0$ on $(- 1, 1)$ if $\int_{- 1}^{1} a u^{'} \overline{v^{'}} d x = 0$ for every $v \in C_{c}^{\infty} (- 1, 1)$ . (Local weak solutions of a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F2]

The coefficient $a$ is measurable and bounded with $1 \leq a \leq 2$ , so $∣ a^{11} ∣ \leq M_{a} = 2$ , and $Re (a^{11} ξ \overline{ξ}) = a ∣ ξ ∣^{2} \geq ∣ ξ ∣^{2}$ for all $ξ \in C$ ; hence $L$ is uniformly elliptic with $θ = 1$ , $M_{a} = 2$ , $M_{b} = M_{c} = 0$ . (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F3]

The Heaviside class $H = 1_{(0, \infty)}$ on $I = (- 1, 1)$ has no weak derivative in $L_{l o c}^{1} (I)$ : if $v \in L_{l o c}^{1} (I)$ satisfied the weak-derivative identity $\int_{I} H φ^{'} d x = - \int_{I} v φ d x$ for every $φ \in C_{c}^{\infty} (I)$ , then the fundamental theorem of calculus would give $\int_{I} v φ d x = φ (0)$ for every test $φ$ , whereas the shrinking bumps $φ_{ϵ} (x) = η (x / ϵ)$ (with $η$ the published smooth bump equal to $1$ on $[- 1 / 2, 1 / 2]$ and supported in $(- 1, 1)$ ) satisfy $φ_{ϵ} (0) = 1$ and $∣ \int_{I} v φ_{ϵ} d x ∣ \leq \int_{[- ϵ, ϵ]} ∣ v ∣ d x \to 0$ as $ϵ ↓ 0$ by absolute continuity of the integral; this contradiction shows that no locally integrable function represents the distributional derivative, which is the Dirac mass at $0$ on $I$ . (Weak derivative of a locally integrable function, Absolute continuity of the integral, A smooth bump between concentric Euclidean balls)

[F4]

The weak derivative is characterized by $\int u φ^{'} = - \int u^{'} φ$ for every compactly supported smooth test. Integrating the displayed piecewise formula for $u$ by parts on $(- 1, 0)$ and $(0, 1)$ gives $u^{'} = 1$ on $(- 1, 0)$ and $u^{'} = 1 / 2$ on $(0, 1)$ , with no point mass because $u$ is continuous at $0$ . Weak differentiation is linear in the class: if a class has a weak derivative in $L_{l o c}^{1}$ , every linear combination with constant coefficients has the corresponding linear combination of weak derivatives. (Weak derivative of a locally integrable function)

[F5]

Assume Countable Choice. If $u \in H_{l o c}^{2} (- 1, 1)$ , then its first weak derivative $u^{'}$ has a weak derivative in $L_{l o c}^{2}$ , and that weak derivative is the distributional second derivative of $u$ . (Weak derivative of a locally integrable function, The notation $H^{k}$ and the reserved zero-boundary symbol)

Counterexample

1.1F4givenalgebra

The solution class and its derivative. By [F4], integration by parts on the two half-intervals gives the weak derivative $u^{'} (x) = 1$ for $x < 0$ and $u^{'} (x) = 1 / 2$ for $x > 0$ ; the boundary terms at $0$ cancel because $u$ is continuous there. Both $u$ and $u^{'}$ lie in $L^{2} (- 1, 1)$ , so $u \in H^{1} (- 1, 1)$ , and equivalently $u^{'} = 1 - \frac{1}{2} 1_{(0, \infty)}$ almost everywhere.

2.1F1step 1.1algebra

The weak equation with zero datum. Since $a u^{'} \equiv 1$ by step 1.1, for every $v \in C_{c}^{\infty} (- 1, 1)$ one has $\int_{- 1}^{1} a u^{'} \overline{v^{'}} d x = \int_{- 1}^{1} \overline{v^{'}} d x = 0$ , the last integral vanishing because $v$ is compactly supported. Hence $u$ is a local weak solution of $- (a u^{'})^{'} = 0$ on $(- 1, 1)$ by [F1].

2.2F3F4F5step 1.1algebra

Failure of $H^{2}$ membership. Suppose $u \in H_{l o c}^{2} (- 1, 1)$ ; by [F5] the weak derivative $u^{'}$ then has a weak derivative $w \in L_{l o c}^{2} (- 1, 1) \subseteq L_{l o c}^{1} (- 1, 1)$ . By step 1.1, $1_{(0, \infty)} = 2 (1 - u^{'})$ a.e., so by linearity of weak differentiation [F4] the Heaviside class would have the locally integrable weak derivative $- 2 w$ , contradicting [F3]. Hence $u \notin H^{2} (- 1, 1)$ , and the distributional second derivative of $u$ is the measure $- \frac{1}{2} δ_{0}$ rather than an $L^{2}$ function.

3.1F2F3F4step 2.1step 2.2algebra∎

Sharpness of the $W^{1, \infty}$ hypothesis and the flux. By [F2] the operator is uniformly elliptic with bounded measurable coefficients, and by step 2.1 it has the $H^{1}$ weak solution $u \notin H^{2}$ with datum $0$ ; since $a = 1 + 1_{(0, \infty)}$ , [F3] and linearity of weak differentiation [F4] show that $a$ has no locally integrable weak derivative either, so $a \notin W_{l o c}^{1, \infty} (- 1, 1)$ . Therefore bounded measurability of the coefficients cannot replace the Lipschitz hypothesis $a^{i j} \in W^{1, \infty} (Ω)$ , with a global derivative bound, of Interior $H^{2}$ regularity for divergence-form equations. Moreover $u^{'}$ jumps from $1$ to $1 / 2$ at $0$ while the flux $a u^{'}$ is identically $1$ on both sides: the quantity continuous across the interface is the flux, not the derivative.

Source notes

Hunter's discussion of composite media (printed p. 120) introduces discontinuous coefficients with continuity of the flux across the interface; Teschl's Lemma 10.16 (printed p. 240) assumes $A \in W^{1, \infty}$ and is therefore not available here. The failure is exhibited at the level of the weak derivative: the coefficient commutator of the differentiated equation is a measure rather than an $L^{2}$ function, so the difference-quotient method of the interior theorem stops exactly at the interface.

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Bootstrapping a smooth Poisson problem

Example

Assume the Axiom of Choice (inherited from the embedding theorem used below) together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, let $f \in C^{\infty} (Ω)$ , and let $u \in H^{1} (Ω)$ be a local weak solution of $- Δ u = f$ on $Ω$ in the sense of Local weak solutions of a divergence-form operator; for instance, when $Ω$ is a nonempty bounded open set and also $f \in L^{2} (Ω)$ , $u$ may be the zero-trace weak Dirichlet solution, whose existence and uniqueness under the Axiom of Choice are Existence and uniqueness for the weak Dirichlet Poisson problem. Iterating Interior $H^{k + 2}$ elliptic regularity with the constant coefficients of the Laplacian gives $u \in H_{l o c}^{m} (Ω)$ for every $m$ , hence a representative of class $C^{\infty}$ by Higher-order Sobolev embedding, and for that representative the equation $- Δ u = f$ holds pointwise on $Ω$ . No boundary data and no boundary regularity are used: the smoothness of $f$ alone permits the induction to continue at every order.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open set $Ω \subseteq R^{n}$ ; $f \in C^{\infty} (Ω)$ ; and a local weak solution $u \in H^{1} (Ω)$ of $- Δ u = f$ on $Ω$ .

[F1]

A class $u \in H^{1} (Ω)$ is a local weak solution of $- Δ u = f$ when $a (u, v) = \int_{Ω} f \overline{v} d x$ for every $v \in C_{c}^{\infty} (Ω)$ , with $a$ the form of the divergence-form operator; by the closure definition this is equivalent to the same identity for every bounded $Ω_{2} ⋐ Ω$ and every $v \in H_{0}^{1} (Ω_{2})$ . (Local weak solutions of a divergence-form operator)

[F2]

Assume Countable Choice. The Laplacian $L = - Δ$ has constant coefficients $a^{i j} = δ^{i j}$ , $b = c = 0$ , hence $a^{i j} \in W_{l o c}^{k + 1, \infty} (Ω)$ and $b, c \in W_{l o c}^{k, \infty} (Ω)$ for every $k \geq 0$ with zeroth-order principal bound $M_{0} = 1$ and all positive-order derivative bounds zero; it is uniformly elliptic with $θ = 1$ . (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F3]

Assume Countable Choice. Let $k \geq 0$ , let $L$ have $a^{i j} \in W_{l o c}^{k + 1, \infty} (Ω)$ and $b^{i}, c \in W_{l o c}^{k, \infty} (Ω)$ with all coefficient derivatives through the indicated orders bounded by constants $M_{ℓ}$ , let $f \in H_{l o c}^{k} (Ω)$ , and let $u \in H^{1} (Ω)$ be a local weak solution of $L u = f$ . Then $u \in H_{l o c}^{k + 2} (Ω)$ . (Interior $H^{k + 2}$ elliptic regularity)

[F4]

Assume the Axiom of Choice. For $n \geq 2$ , let $Ω_{0}$ be a bounded extension domain in $R^{n}$ , let $k \geq 1$ , $1 \leq p < \infty$ with $k p > n$ . If $m \geq 0$ and $0 < α < 1$ satisfy $m + α < k - n / p$ , then every $u \in W^{k, p} (Ω_{0})$ has a representative in $C^{m, α} (\overline{Ω_{0}})$ with norm bounded by $C ∥ u ∥_{W^{k, p} (Ω_{0})}$ . In particular, $p = 2$ and an integer $k > n / 2$ give a continuous representative on $Ω_{0}$ . (Higher-order Sobolev embedding)

[F5]

Every open ball in $R^{n}$ , $n \geq 2$ , is a bounded extension domain: it is a bounded $C^{\infty}$ domain in the graph sense, so Bounded C^k domains admit integer-order Sobolev extension supplies an extension operator. For $n = 1$ , on each bounded open interval $I ⋐ Ω$ every class in $W^{1, 2} (I)$ has a unique continuous, locally absolutely continuous representative by One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives. (Sobolev extension domains and extension operators)

[F6]

Assume Countable Choice. If $u \in H_{l o c}^{2} (Ω)$ is a local weak solution of $L u = f$ with $a^{i j} \in W^{1, \infty} (Ω)$ , bounded first coefficient derivatives, and $f \in L_{l o c}^{2} (Ω)$ , then the equation holds pointwise almost everywhere, with $D_{i} (a^{i j} D_{j} u)$ understood through the a.e. defined product $(D_{i} a^{i j}) D_{j} u + a^{i j} D_{i} D_{j} u$ ; for the constant coefficients of the Laplacian this is the a.e. identity $- \sum_{i} D_{i} D_{i} u = f$ . (Interior $H^{2}$ regularity for divergence-form equations)

Verification

1.1F1F2F3algebragiven

Local smoothness at every order. Fix $k \geq 0$ and a bounded open $Ω^{''} ⋐ Ω$ . By [F2] the Laplacian satisfies the coefficient hypotheses of [F3] at order $k$ , and $f \in C^{\infty} (Ω)$ gives $f \in H_{l o c}^{k} (Ω)$ ; since $u$ is a local weak solution, [F3] gives $u \in H_{l o c}^{k + 2} (Ω)$ , hence $u \in H^{k + 2} (Ω^{''})$ . As $k$ was arbitrary, $u \in H_{l o c}^{m} (Ω)$ for every $m \geq 1$ .

2.1F4F5step 1.1algebra

A smooth representative. If $n \geq 2$ , fix a ball $B ⋐ Ω$ and an integer $m \geq 0$ . Choose a slightly larger ball $B^{'} ⋐ Ω$ and an integer $s > m + n / 2$ . Step 1.1 gives $u \in H^{s} (B^{'})$ , and [F5] makes $B^{'}$ a bounded extension domain; [F4] then gives a $C^{m, α}$ representative on $\overline{B^{'}}$ for some $α > 0$ . For different $m$ these representatives agree everywhere on overlaps: they are continuous and represent the same almost-everywhere class. Thus they define a $C^{\infty}$ representative on $B$ . If $n = 1$ , fix a bounded open interval $I ⋐ Ω$ . Step 1.1 gives $D^{j} u \in H^{1} (I)$ for every $j \geq 0$ . By [F5] each has a unique continuous, locally absolutely continuous representative $g_{j}$ . The weak derivative of $g_{j}$ is $D^{j + 1} u$ , so the fundamental theorem gives $g_{j} (y) - g_{j} (x) = \int_{x}^{y} g_{j + 1} (t) d t$ for $x < y$ in $I$ ; continuity of $g_{j + 1}$ implies $g_{j} \in C^{1} (I)$ and $g_{j}^{'} = g_{j + 1}$ . Iterating, $g_{0}$ is $C^{\infty} (I)$ . These local representatives agree on overlaps, yielding a smooth representative on all components of $Ω$ .

3.1F6step 2.1algebragiven∎

The equation holds pointwise. For the $C^{\infty}$ representative of step 2.1, $u \in H_{l o c}^{2} (Ω)$ and $f \in L_{l o c}^{2} (Ω)$ , so by [F6] the equation $- Δ u = f$ holds pointwise almost everywhere, the Laplacian's expression being the a.e. function $- \sum_{i} D_{i} D_{i} u$ . Both sides, $- \sum_{i} D_{i} D_{i} u$ by step 2.1 and $f$ by hypothesis, are continuous on $Ω$ , and two continuous functions that agree almost everywhere on an open set agree at every point. Thus the smoothed representative solves the classical equation pointwise.

Source notes

Hunter's Corollary 4.29 and Laugesen's Theorems 5.8-5.9 (printed pp. 114 and 111-112, read in full) iterate the interior estimate to obtain $H_{l o c}^{m}$ for every $m$ and then a smooth representative; the same two-step pattern is used above. The example separates the two inputs: the coefficient regime of the Laplacian never obstructs the induction, and the smoothness of $f$ is exactly what allows the data order $k$ to increase; the Axiom of Choice is carried only by the Sobolev embedding used for the representative.

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Higher elliptic regularity cannot gain more than two derivatives

Statement refuted

Assume Countable Choice. For every integer $k \geq 0$ , smooth constant coefficients and $f \in H_{l o c}^{k} (Ω)$ force every $H^{1}$ local weak solution of $- u^{''} = f$ into $H_{l o c}^{k + 3} (Ω)$ .

Facts & Assumptions

Given: Countable Choice; a fixed integer $k \geq 0$ ; the interval $Ω = (- 1, 1)$ ; the datum $f (x) = ∣ x ∣^{k + 1 / 2}$ ; and the function $u (x) = - A ∣ x ∣^{k + 5 / 2}$ with $A = ((k + \frac{5}{2}) (k + \frac{3}{2}))^{- 1} > 0$ .

[F1]

On each half-interval the classical derivative of order $j$ of $f (x) = ∣ x ∣^{k + 1 / 2}$ is a nonzero constant times $∣ x ∣^{k + 1 / 2 - j}$ , with a possible sign change across $0$ . For $j \leq k$ these derivatives tend to $0$ at $0$ and are in $L^{2}$ ; the order $k + 1$ derivative is locally integrable with magnitude a positive constant times $∣ x ∣^{- 1 / 2}$ , but is not in $L^{2}$ near $0$ . Since all derivatives through order $k$ extend continuously across $0$ , the piecewise classical derivatives are the weak derivatives through order $k + 1$ , with no point-mass terms. Thus $f \in H^{k} ((- 1, 1))$ but $f \notin H_{l o c}^{k + 1} ((- 1, 1))$ near $0$ . (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function)

[F2]

A class $u \in H^{1} ((- 1, 1))$ is a local weak solution of $- u^{''} = f$ if $\int_{- 1}^{1} u^{'} \overline{φ^{'}} d x = \int_{- 1}^{1} f \overline{φ} d x$ for every $φ \in C_{c}^{\infty} ((- 1, 1))$ . (Local weak solutions of a divergence-form operator)

[F3]

Put $p = k + \frac{5}{2}$ and $A = (p (p - 1))^{- 1}$ . For $u = - A ∣ x ∣^{p}$ , its derivatives through order $k + 2$ are piecewise constant multiples of $∣ x ∣^{p - j}$ , hence lie in $L^{2} ((- 1, 1))$ ; the $(k + 3)$ rd derivative has magnitude a positive constant times $∣ x ∣^{- 1 / 2}$ and is not locally in $L^{2}$ at $0$ . The derivatives through order $k + 2$ extend continuously across $0$ , so these piecewise formulas are the weak derivatives and no delta mass occurs. (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function)

[F4]

Assume Countable Choice. For the operator $L = - u^{''}$ in dimension one, $a^{11} = 1$ and $b = c = 0$ , so it is uniformly elliptic with $θ = M_{a} = 1$ , $M_{b} = M_{c} = 0$ , and its constant coefficients lie in $W_{l o c}^{k + 1, \infty}$ for every $k$ . If $f \in H_{l o c}^{k} (Ω)$ and $u \in H^{1} (Ω)$ is a local weak solution of $L u = f$ , the interior theorem gives $u \in H_{l o c}^{k + 2} (Ω)$ . (Interior $H^{k + 2}$ elliptic regularity, Uniformly elliptic divergence-form operators and their sesquilinear forms, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Counterexample

1.1F1algebragiven

Data regularity. The piecewise derivative calculation in [F1] shows $f \in H^{k} ((- 1, 1))$ but $f \notin H_{l o c}^{k + 1}$ near $0$ .

1.2F3algebragiven

The solution's regularity. With $u = - A ∣ x ∣^{p}$ from [F3], one has $- u^{''} = A p (p - 1) ∣ x ∣^{p - 2} = ∣ x ∣^{k + 1 / 2} = f$ on both sides of $0$ . Since $u^{'}$ is continuous at $0$ and $u^{''} = - f$ is locally integrable, this also holds distributionally across $0$ . The piecewise derivative calculation in [F3] gives $u \in H^{k + 2} ((- 1, 1))$ but $u \notin H_{l o c}^{k + 3}$ near $0$ .

2.1F2step 1.2algebra

The weak equation. For every $φ \in C_{c}^{\infty} ((- 1, 1))$ , the distributional identity $- u^{''} = f$ gives $\int_{- 1}^{1} u^{'} \overline{φ^{'}} d x = \int_{- 1}^{1} f \overline{φ} d x$ . Thus $u$ is a local weak solution by [F2].

3.1F4step 1.1step 2.1algebra

The interior $H^{k + 2}$ theorem agrees with the direct calculation. The Laplacian has smooth constant coefficients, $f \in H_{l o c}^{k}$ , and the weak solution satisfies the hypotheses of [F4], so that theorem gives $u \in H_{l o c}^{k + 2}$ . The explicit formula in step 1.2 gives the stronger global $H^{k + 2}$ membership.

4.1F1step 1.1step 3.1algebra∎

No third extra local derivative. If $u \in H^{k + 3} (I)$ on a neighborhood $I$ of $0$ , then its third weak derivative lies in $H^{k} (I)$ . But $u^{'''} = - f^{'}$ in distributions, so $f^{'} \in H^{k} (I)$ and hence $f \in H^{k + 1} (I)$ , contradicting step 1.1. Therefore this interior weak solution belongs to $H_{l o c}^{k + 2}$ but not to $H_{l o c}^{k + 3}$ near $0$ , even with constant smooth coefficients.

Source notes

Hunter's Theorems 4.28 and 4.31 and Teschl's Corollary 10.19 (printed pp. 114-116 and p. 243) record the gain of exactly two derivatives. The interior cusp $f = ∣ x ∣^{k + 1 / 2}$ shows the sharpness at an interior point, rather than only at the boundary; the exact $H^{k + 2}$ and failure of $H_{l o c}^{k + 3}$ follow directly from the explicit formula.

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The reentrant sector singularity has an explicit Sobolev threshold

Example

Assume Countable Choice. Let $π < ω < 2 π$ , let $S_{ω} = {(r \cos θ, r \sin θ) : 0 < r < 1, 0 < θ < ω}$ be the reentrant sector, let $α = π / ω \in (0, 1)$ , and let $u (r, θ) = r^{α} \sin (α θ)$ . Then $Δ u = 0$ in $S_{ω}$ , $u$ vanishes on the two radial edges $θ = 0$ and $θ = ω$ , and for every integer $m \geq 0$ $u \in H^{m} (S_{ω}) ⟺ m < 1 + α = 1 + \frac{π}{ω};$ in particular $u \in H^{1} (S_{ω}) ∖ H^{2} (S_{ω})$ and each additional whole derivative beyond $H^{1}$ is unavailable exactly by the deficit $1 - π / ω$ . For noninteger $s = m + t$ with integer $m \geq 0$ and $0 < t < 1$ , define the intrinsic Slobodeckij scale here by requiring $u \in H^{m} (S_{ω})$ and finite seminorm $\int_{S_{ω}} \int_{S_{ω}} \frac{∣ D^{β} u (x) - D^{β} u (y) ∣^{2}}{∣ x - y ∣^{2 + 2 t}} d x d y$ for each weak derivative with $∣ β ∣ = m$ . With this convention, for every real $s \geq 0$ , $u \in H^{s} (S_{ω}) ⟺ s < 1 + α = 1 + \frac{π}{ω}$ . The integer threshold follows from polar-coordinate integrals; the fractional threshold follows from a dyadic-shell estimate and a matching scaled-pair lower bound.

Facts & Assumptions

Given: $ω \in (π, 2 π)$ , the sector $S_{ω}$ above, $α = π / ω \in (0, 1)$ , and $u (r, θ) = r^{α} \sin (α θ)$ .

[F1]

A class in $H^{1} (S_{ω})$ is a local weak solution of $- Δ u = 0$ on $S_{ω}$ if $\int_{S_{ω}} \nabla u \cdot \overline{\nabla v} d x = 0$ for every $v \in C_{c}^{\infty} (S_{ω})$ . (Local weak solutions of a divergence-form operator)

[F2]

For every $C^{2}$ function $w$ on the punctured plane, the chain rule gives $Δ w = w_{r r} + r^{- 1} w_{r} + r^{- 2} w_{θ θ}$ in polar coordinates; in particular $Δ (r^{α} \sin (α θ)) = 0$ on the sector. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Weak derivative of a locally integrable function)

[F3]

Polar integration on the sector is $\int_{S_{ω}} f d x = \int_{0}^{ω} \int_{0}^{1} f (r \cos θ, r \sin θ) r d r d θ$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F4]

A class $u$ lies in $H^{m} (S_{ω})$ for an integer $m \geq 0$ exactly when all its weak partial derivatives of order $\leq m$ lie in $L^{2} (S_{ω})$ ; the weak derivatives of the smooth function $u$ on $S_{ω} ∖ {0}$ are the classical ones, and on the sector $r > 0$ the classical derivatives $D^{j} u$ are bounded by constant multiples of $r^{α - j}$ . Along each fixed ray $x = r e_{r}$ , the radial derivative satisfies $\partial_{r}^{m} u (r, θ) = D^{m} u (r e_{r}) [e_{r}, \dots, e_{r}]$ ; since $∣ e_{r} ∣ = 1$ , if all Cartesian derivatives of order $m$ lie in $L^{2}$ , then this radial derivative also lies in $L^{2}$ . (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function)

[F5]

The reentrant sector is a bounded Lipschitz domain that fails the $C^{1}$ boundary-chart condition at its vertex, and $u$ is a local weak solution of $- Δ u = 0$ on it with $u \in H^{1} (S_{ω}) ∖ H^{2} (S_{ω})$ . (Boundary $H^{2}$ regularity needs domain regularity)

[F6]

Let $w (r, θ) = r^{γ} Ψ (θ)$ on this sector, where $γ > - 1$ and $Ψ$ is smooth on $[0, ω]$ . For $0 < t < 1$ , the intrinsic seminorm $\int_{S_{ω}} \int_{S_{ω}} \frac{∣ w (x) - w (y) ∣^{2}}{∣ x - y ∣^{2 + 2 t}} d x d y$ is finite if $t < γ + 1$ . To see this, put $A_{j} = {2^{- j - 1} < r < 2^{- j}, 0 < θ < ω}$ and $λ_{j} = 2^{- j}$ . The angular formula extends smoothly to a slightly larger interval because $ω < 2 π$ , so near pairs in comparable shells satisfy $∣ w (x) - w (y) ∣ \leq C λ_{j}^{γ - 1} ∣ x - y ∣$ ; integrating such pairs gives $C λ_{j}^{2 γ + 2 - 2 t}$ . Separated pairs in comparable shells give the same bound from $∣ w ∣ \leq C λ_{j}^{γ}$ and $∣ x - y ∣ \geq c λ_{j}$ . For noncomparable shells $A_{j}, A_{k}$ with $k \geq j + 2$ , $∣ x - y ∣ \geq c λ_{j}$ ; integrating the two terms $λ_{j}^{2 γ}$ and $λ_{k}^{2 γ}$ in $∣ w (x) - w (y) ∣^{2}$ and summing over $k$ gives at most $C λ_{j}^{2 γ + 2 - 2 t}$ , with the inner sum geometric since $γ > - 1$ . The final sum over $j$ converges exactly when $t < γ + 1$ .

[F7]

For $D u = \nabla u = α r^{α - 1} (\sin ((α - 1) θ), \cos ((α - 1) θ))$ , choose two small disjoint balls $A, B$ compactly contained in ${1 / 2 < r < 1, 0 < θ < ω}$ , centered at the same radius and at angles $ω / 4$ and $3 ω / 4$ . Their gradient values differ because the direction-angle difference is $(α - 1) ω / 2 = (π - ω) / 2 \neq 0$ ; shrinking the balls gives $∣ D u (x) - D u (y) ∣ \geq c > 0$ on $A \times B$ . By homogeneity $D u (λ x) = λ^{α - 1} D u (x)$ , the order- $t$ seminorm integral over $2^{- j} A \times 2^{- j} B$ is at least $c^{'} 2^{- j (2 α - 2 t)}$ . These product sets are pairwise disjoint as $j$ varies, so their sum diverges for $t \geq α$ .

Verification

1.1F2algebragiven

Harmonicity and edge vanishing. By [F2] the polar Laplacian of $w = r^{α} \sin (α θ)$ is $(α (α - 1) + α - α^{2}) r^{α - 2} \sin (α θ) = 0$ , so $u$ is harmonic and $C^{\infty}$ on $S_{ω}$ ; and $\sin (α \cdot 0) = \sin (π) = 0 = \sin (α ω)$ shows that $u$ vanishes on both radial edges.

1.2F3F4algebra

Membership below the threshold. Since $∣ D^{j} u (r, θ) ∣ \leq C_{j} r^{α - j}$ for the classical derivatives by [F4], [F3] gives $\int_{S_{ω}} ∣ D^{j} u ∣^{2} d x \leq C_{j}^{2} \int_{0}^{ω} \int_{0}^{1} r^{2 α - 2 j} r d r d θ = C_{j}^{'} ⁣ \int_{0}^{1} r^{2 α - 2 j + 1} d r$ , which is finite whenever $2 α - 2 j + 1 > - 1$ , that is $j < α + 1$ . Hence every weak derivative of order $j \leq m$ lies in $L^{2} (S_{ω})$ whenever the integer $m$ satisfies $m < 1 + α$ , and then $u \in H^{m} (S_{ω})$ by [F4].

1.3F3F4algebra

Non-membership at and above the threshold. Let $m \geq α + 1$ be an integer, so $m \geq 1$ because $α < 1$ , and put $c_{m} : = α (α - 1) \dots (α - m + 1) \neq 0$ . Differentiating along a fixed ray gives $\partial_{r}^{m} u = c_{m} r^{α - m} \sin (α θ)$ . Since $2 α - 2 m + 1 \leq - 1$ and $\int_{0}^{ω} \sin^{2} (α θ) d θ > 0$ , [F3] gives $\int_{S_{ω}} ∣ \partial_{r}^{m} u ∣^{2} d x = c_{m}^{2} (\int_{0}^{ω} \sin^{2} (α θ) d θ) (\int_{0}^{1} r^{2 α - 2 m + 1} d r) = + \infty .$ By [F4], membership in $H^{m}$ would force this radial derivative to lie in $L^{2}$ , so $u \notin H^{m} (S_{ω})$ .

2.1step 1.2step 1.3

The integer threshold. Steps 1.2 and 1.3 give, for every integer $m \geq 0$ , $u \in H^{m} (S_{ω}) ⟺ m < 1 + α .$

2.2F1F5step 1.2algebra

The stated particular cases. For $m = 1$ the criterion gives $1 < 1 + α$ because $α > 0$ , so $u \in H^{1} (S_{ω})$ ; for $m = 2$ it gives $2 < 1 + α$ , which fails because $α < 1$ and $ω > π$ ; hence $u \notin H^{2} (S_{ω})$ . These conclusions agree with the local weak-solution statement of [F5], which records the same function as the reentrant-corner witness and with [F1]'s definition of a local weak solution.

2.3F6step 1.2algebra

Fractional membership below the threshold. Let $s = m + t$ be noninteger with $m = 0$ or $m = 1$ and $0 < t < 1$ . If $m = 0$ , then $u \in L^{2}$ by step 1.2 and [F6] applies with $γ = α$ , giving finite $H^{t}$ seminorm since $t < 1 < 1 + α$ . If $m = 1$ , then $u \in H^{1}$ and each component of $D u$ has the form in [F6] with degree $γ = α - 1$ ; its $H^{t}$ seminorm is finite when $t < γ + 1 = α$ , exactly when $s = 1 + t < 1 + α$ .

3.1F7step 2.1step 2.2algebra∎

Fractional nonmembership and all higher orders. For $1 + α \leq s < 2$ , put $t = s - 1 \geq α$ . By [F7], $D u$ has infinite order- $t$ seminorm, so the defining condition for $H^{s}$ fails. For $s \geq 2$ , membership in the defined real-order scale entails membership in $H^{2} (S_{ω})$ , which step 2.2 rules out; $s = 0$ is covered by $u \in L^{2}$ . Together with step 2.3 and the integer criterion of step 2.1, this proves $u \in H^{s} (S_{ω})$ exactly when $s < 1 + α$ .

Scope note

The real-order statement uses the intrinsic Slobodeckij convention specified in the statement; this fixes the fractional scale on the reentrant sector and does not rely on an unstated extension or boundary regularity theorem.

Source notes

Teschl's Example 10.1 (printed p. 242) is the source for the harmonic model function and the failure of $H^{2}$ in a reentrant sector. The quantitative threshold for integer orders follows from the explicit $r^{α - j}$ derivative bounds, the polar integral, and the directional-derivative bound in [F4].

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Smooth interior data do not repair incompatible Dirichlet corner values

Statement refuted

On the square $Ω = (0, 1)^{2}$ , every boundary datum whose restriction to each open side is smooth is the boundary trace of a function $u \in C (\overline{Ω})$ harmonic in $Ω$ , that is, satisfying $- Δ u = 0$ there.

Facts & Assumptions

Given: The Axiom of Choice; the square $Ω = (0, 1)^{2}$ ; and the boundary datum $g (x, 0) = 1, g (0, y) = 0, g (1, y) = 0, g (x, 1) = 0 (0 < x, y < 1),$ each of the four functions being constant and therefore smooth on its open side.

[F1]

The refuted assertion concerns the Laplace Dirichlet problem $- Δ u = 0$ in $Ω$ with the prescribed sidewise boundary values; this is the uniformly elliptic divergence-form convention with $a^{i j} = δ^{i j}$ and $b = c = 0$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, Weak Dirichlet solutions for a divergence-form operator).

[F2]

Assume the Axiom of Choice. The square $Q = (0, 1)^{2}$ is an $H^{2}$ extension domain by explicit reflection. For $h \in H^{2} (0, 1)$ extend across $0$ by $3 h (- x) - 2 h (- 2 x)$ on $(- 1 / 2, 0)$ and across $1$ by $3 h (2 - x) - 2 h (3 - 2 x)$ on $(1, 3 / 2)$ , retaining $h$ on $[0, 1]$ . For an $H^{2}$ interval class, the opened one-dimensional representative corollary applied to $h$ and $h^{'}$ supplies continuous endpoint values. These formulas match the value and first derivative at each join, and affine changes of variables bound the $H^{2}$ norm on the enlarged interval. To apply the formula to $u \in H^{2} (Q)$ , Fubini and the weak-derivative identities tested against products of one-dimensional smooth tests show that almost every coordinate slice of $u$ is $H^{2}$ , and that the slices of its transverse first derivative are $H^{1}$ . One can choose a common null set by using a countable dense family of interval tests; passage to any test follows by the L2 bounds. On these slices the reflected formulas match the function and its normal first derivative, so integration by parts on the joined intervals has no interface terms. Transverse weak derivatives commute with the reflection by affine change of variables in the tensor test identities; for the mixed derivative only the H1 matching of the transverse first-derivative slices is needed. Thus each coordinate operation bounds all pure and mixed weak derivatives through order two. Applying them successively gives a bounded extension from $H^{2} (Q)$ to $H^{2} ((- 1 / 2, 3 / 2)^{2})$ ; multiplying by a smooth cutoff equal to $1$ on $\overset{ˉ}{Q}$ and supported in the larger rectangle, then extending by zero, gives an $H^{2} (R^{2})$ extension. Thus $Q$ is a bounded extension domain. Since $2 \cdot 2 > 2$ , Higher-order Sobolev embedding gives a continuous representative on $\overset{ˉ}{Q}$ for every $H^{2} (Q)$ class. (Sobolev extension domains and extension operators, A Euclidean bump for a compact set inside an open set, Fubini's theorem for L^1 functions on a sigma-finite product)

[F3]

On a bounded $C^{2}$ domain the global $H^{2}$ estimate is available for the zero-trace problem; it presupposes a compatible datum and does not by itself produce one. (Global $H^{2}$ Dirichlet regularity, Regularity estimates do not create boundary compatibility)

[F4]

The square is bounded Lipschitz but is not $C^{1}$ at its four corners: the two incident straight edges do not form a single $C^{1}$ boundary graph. Thus the bounded $C^{2}$ hypothesis of the global boundary theorem does not apply to this domain. (Bounded C^k domains and boundary charts)

[F5]

Two continuous functions on $Q$ that agree almost everywhere agree everywhere, since a nonzero difference at one point remains nonzero on an open ball of positive measure. A continuous representative on $\overset{ˉ}{Q}$ that attains the prescribed constant values on the open sides must therefore have equal limits along the two sides at each shared corner.

Counterexample

1.1F1algebragiven

No solution continuous on the closure. Suppose $u \in C (\overline{Ω})$ has boundary trace agreeing with $g$ on each open side. Continuity of $u$ at the corner $(0, 0)$ makes the limits of $u$ along the two sides through the corner equal: taking $(x, 0) \to (0, 0)$ with $x ↓ 0$ gives $u (x, 0) = g (x, 0) = 1 \to 1$ , while taking $(0, y) \to (0, 0)$ with $y ↓ 0$ gives $u (0, y) = g (0, y) = 0 \to 0$ . Since $1 \neq 0$ , no such continuous solution exists, for any divergence-form operator; in particular the refuted statement fails on the square with this datum.

2.1F2F5step 1.1algebra

No $H^{2}$ representative can realize the sidewise data. Suppose a class $u \in H^{2} (Q)$ had a continuous representative $h$ on $\overset{ˉ}{Q}$ whose restrictions to the open sides are the prescribed data. By [F2] the same Sobolev class has a representative $\tilde{u} \in C (\overset{ˉ}{Q})$ . They agree almost everywhere on $Q$ , so [F5] makes them equal everywhere on $Q$ ; continuity then makes them equal on $\overset{ˉ}{Q}$ . Thus $\tilde{u}$ has the same side values as $h$ , and step 1.1 gives a contradiction. Hence no $H^{2}$ class has a continuous representative realizing these sidewise data, and a fortiori no smoother classical solution does.

3.1F1F3F4step 2.1algebra∎

Compatibility is not created by regularity. The interior datum is $f = 0$ for $- Δ u = 0$ , but the square has corners and is not a $C^{2}$ domain by [F4]. The prescribed boundary values are incompatible with any continuous representative by step 1.1; a regularity estimate cannot create the missing boundary compatibility, as [F3] records. Thus smooth sidewise boundary data and coefficients do not repair corner incompatibility.

Source notes

Hunter's Theorems 4.30-4.31 (printed pp. 114-116) are stated for zero-trace classes, so the inhomogeneous datum must first be lifted; Simon's Lecture 9 Theorem 1 (printed pp. 88-90) likewise assumes a localized zero-Dirichlet class; it is not a source for the square's incompatible sidewise data. The two-limit contradiction at the corner is elementary and uses only continuity; the $H^{2}$ clause additionally uses the Sobolev embedding of [F2], which is why this item states the Axiom of Choice even though the primary refutation of continuous solutions needs none.

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The $H^{2}$ estimate needs the $L^{2}$ kernel term without injectivity

Statement refuted

For every bounded $C^{2}$ domain $Ω \subset R^{n}$ , $n \geq 2$ , and every uniformly elliptic divergence-form operator $L$ with bounded coefficients, the estimate $∥ u ∥_{H^{2} (Ω)} \leq C ∥ L u ∥_{L^{2} (Ω)}$ holds for every weak solution $u \in H_{0}^{1} (Ω)$ of the homogeneous Dirichlet problem, with $C$ depending only on the operator and domain data, even when the homogeneous Dirichlet operator has a nontrivial kernel.

Facts & Assumptions

Given: Countable Choice; $Ω = B_{1} (0) \subset R^{2}$ , $a (r) = \frac{1}{4} - \frac{1}{8} r^{2}$ , $A (x) = a (∣ x ∣) I$ , $L w = - div (A \nabla w) - w$ , zero datum $f = 0$ , and $u (x) = 1 - ∣ x ∣^{2}$ .

[F1]

A weak Dirichlet solution is a function $w \in H_{0}^{1} (B_{1})$ satisfying the form identity $\int_{B_{1}} (A \nabla w \cdot \overline{\nabla v} - w \overline{v}) d x = \int_{B_{1}} f \overline{v} d x$ for all $v \in H_{0}^{1} (B_{1})$ (Weak Dirichlet solutions for a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F2]

For $0 \leq r \leq 1$ , $\frac{1}{8} \leq a (r) \leq \frac{1}{4}$ . Thus $A$ is smooth, bounded, and uniformly elliptic with ellipticity constant $θ = 1 / 8$ ; the lower-order coefficients are bounded, with $b = 0$ and $c = - 1$ (Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

The function $u = 1 - ∣ x ∣^{2}$ is smooth on $\overline{B_{1}}$ , zero on $\partial B_{1}$ , and nonzero, so $u \in H^{2} (B_{1})$ . To prove $u \in H_{0}^{1} (B_{1})$ directly, choose smooth radial cutoffs $η_{ε}$ equal to one for $r \leq 1 - 2 ε$ and zero for $r \geq 1 - ε$ , with $∣ D η_{ε} ∣ \leq C / ε$ , using a rescaled fixed smooth step. Then $η_{ε} u \in C_{c}^{\infty} (B_{1})$ ; on the boundary strip $∣ u ∣ \leq C ε$ , $∣ D u ∣ \leq 2$ , and its area is at most $C ε$ . Thus $∥ η_{ε} u - u ∥_{H^{1}}^{2} \leq C ε \to 0$ . Consequently $u \in H^{2} (B_{1}) \cap H_{0}^{1} (B_{1})$ and $∥ u ∥_{H^{2} (B_{1})} > 0$ . With $r = ∣ x ∣$ , $\nabla u = - 2 x$ and $a^{'} (r) = - r / 4$ , whence $- div (a (r) \nabla u) = 4 a (r) + 2 r a^{'} (r) = 1 - r^{2} = u .$ Therefore $L u = 0$ pointwise.

[F4]

On bounded $C^{2}$ domains in dimensions $n \geq 2$ , the global theorem Global $H^{2}$ Dirichlet regularity includes an $L^{2}$ term on the right, while the estimate without that term is supplied under a trivial-kernel hypothesis by The global $H^{2}$ estimate without the $L^{2}$ term under uniqueness.

Counterexample

1.1givenF2

The coefficient and domain assumptions hold. The disk $B_{1} \subset R^{2}$ is a bounded smooth domain, and [F2] verifies uniform ellipticity and bounded coefficients.

1.2F3

The function is an admissible nonzero zero-boundary element. By [F3], $u \in H^{2} (B_{1}) \cap H_{0}^{1} (B_{1})$ and $∥ u ∥_{H^{2} (B_{1})} > 0$ .

1.3F1F3

It is a weak homogeneous solution. Since $L u = 0$ pointwise, integration by parts first gives the weak form identity for $C_{c}^{\infty} (B_{1})$ tests. The form is continuous on $H_{0}^{1} (B_{1})$ , so density extends the identity to all such tests. Thus $u$ is a nonzero weak Dirichlet solution with datum zero.

2.1step 1.3F3F4algebra∎

The estimate without the $L^{2}$ term fails. For every finite $C$ , its right side is $C ∥ L u ∥_{L^{2} (B_{1})} = 0$ , while the left side is strictly positive by [F3]. Hence no estimate of this form holds without a kernel condition, as reflected in [F4].

Source notes

Laugesen's note after Theorem 5.10 (printed p. 113) gives a one-dimensional kernel example for the same general obstruction; Hunter's spectral discussion (printed p. 110) describes the corresponding zero-eigenvalue alternative. The disk example above is verified directly and does not invoke a spectral theorem.

Interior and Boundary Sobolev Elliptic Regularity — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Poisson's equation with L2L2 data gains two interior derivatives

Example

Facts & Assumptions

Verification

Source notes

A piecewise-smooth coefficient gives an H2H2 solution that is not twice classically differentiable

Example

Facts & Assumptions

Verification

Source notes

Interior regularity does not imply boundary regularity

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Boundary H2H2 regularity needs domain regularity

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Bounded discontinuous elliptic coefficients need not give H2H2 solutions

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Bootstrapping a smooth Poisson problem

Example

Facts & Assumptions

Verification

Source notes

Higher elliptic regularity cannot gain more than two derivatives

Statement refuted

Facts & Assumptions

Counterexample

Source notes

The reentrant sector singularity has an explicit Sobolev threshold

Example

Facts & Assumptions

Verification

Scope note

Source notes

Smooth interior data do not repair incompatible Dirichlet corner values

Statement refuted

Facts & Assumptions

Counterexample

Source notes

The H2H2 estimate needs the L2L2 kernel term without injectivity

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Sources

Poisson's equation with $L^{2}$ data gains two interior derivatives

A piecewise-smooth coefficient gives an $H^{2}$ solution that is not twice classically differentiable

Boundary $H^{2}$ regularity needs domain regularity

Bounded discontinuous elliptic coefficients need not give $H^{2}$ solutions

The $H^{2}$ estimate needs the $L^{2}$ kernel term without injectivity