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Schauder and $L^{p}$ Elliptic Estimates — Examples

1 · Prerequisites

2 · Summary

The examples check the radius bookkeeping of the Schauder estimate, exhibit the logarithmic loss that makes the endpoint $α = 1$ fail, record the non-Dini and jump-coefficient obstructions to Schauder regularity, separate weak boundary regularity on Lipschitz domains from the a priori estimates, and reduce the method of continuity to a one-dimensional eigenvalue model.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

The Schauder estimate on a quadratic Poisson solution: radius powers balance

Example

Assume Countable Choice when invoking the estimate supplier for $n \geq 2$ . Let $n \geq 1$ , $0 < α < 1$ , $c \neq 0$ , $R > 0$ , $x_{0} \in R^{n}$ and $u (x) : = \frac{c (R^{2} - ∣ x - x_{0} ∣^{2})}{2 n}$ . Then $- Δ u \equiv c$ on $B_{R} (x_{0})$ and $u = 0$ on $\partial B_{R} (x_{0})$ . On the inner ball $B_{R / 2} (x_{0})$ one computes exactly $\sup_{B_{R / 2}} ∣ u ∣ = \frac{∣ c ∣ R^{2}}{2 n}, \max_{∣ β ∣ = 1} \sup_{B_{R / 2}} ∣ D^{β} u ∣ = \frac{∣ c ∣ R}{2 n}, ∣ D^{2} u ∣ \equiv \frac{∣ c ∣}{n}, [D^{2} u]_{0, α; B_{R / 2}} = 0,$ so that $∥ u ∥_{2, α; B_{R / 2} (x_{0})}^{*} = \frac{∣ c ∣ R^{2}}{n}, ∥ u ∥_{\infty; B_{R}} + R^{2} ∥ L u ∥_{\infty; B_{R}} + R^{2 + α} [L u]_{0, α; B_{R}} = ∣ c ∣ R^{2} (\frac{1}{2 n} + 1),$ for the operator $L = Δ$ (so that $L u = Δ u = - c$ ). Both sides are proportional to $∣ c ∣ R^{2}$ with constants independent of $R$ : the radius powers balance exactly. The example also verifies the dilation identity $∥ u ∥_{2, α; B_{R / 2} (x_{0})}^{*} = ∥ u (x_{0} + R \cdot) ∥_{2, α; B_{1 / 2} (0)}^{*}$ of Hölder spaces $C^{k, α}$ , closure and interior scaled norms, and $C^{k, α}$ domains.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < α < 1$ , $c \neq 0$ , $R > 0$ , $x_{0} \in R^{n}$ , the quadratic $u (x) = c (R^{2} - ∣ x - x_{0} ∣^{2}) / (2 n)$ , and the operator $L = Δ$ in the nondivergence convention in which the estimate is stated with data $L u$ .

[F1]

The Laplacian is $Δ = \sum_{i} \partial_{i} \partial_{i}$ and the scaled interior norm is $∥ w ∥_{2, α; B_{ρ} (x_{0})}^{*} = \sum_{j = 0}^{2} ρ^{j} \max_{∣ β ∣ = j} \sup_{B_{ρ}} ∣ D^{β} w ∣ + ρ^{2 + α} \max_{∣ β ∣ = 2} [D^{β} w]_{0, α; B_{ρ}}$ , with $[w]_{0, α; B} = \sup_{x \neq y \in B} ∣ w (x) - w (y) ∣ / ∣ x - y ∣^{α}$ and the same formula on balls of radius $R$ ; under $v (z) = w (x_{0} + R z)$ one has $∥ v ∥_{2, α; B_{1} (0)}^{*} = ∥ w ∥_{2, α; B_{R} (x_{0})}^{*}$ . (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field, Hölder spaces $C^{k, α}$ , closure and interior scaled norms, and $C^{k, α}$ domains, Local Hölder and scaled C-two-alpha norms on balls)

[F2]

For $n \geq 2$ , the interior Schauder estimate for $Δ$ (Interior Schauder estimate for uniformly elliptic equations): if $w \in C^{2} (B_{R} (x_{0})) \cap L^{\infty} (B_{R} (x_{0}))$ satisfies $Δ w = g$ pointwise with $g \in C^{0, α} (B_{R} (x_{0}))$ , then $∥ w ∥_{2, α; B_{R / 2} (x_{0})}^{*} \leq C_{n, α} (∥ w ∥_{\infty; B_{R}} + R^{2} ∥ g ∥_{\infty; B_{R}} + R^{2 + α} [g]_{0, α; B_{R}})$ ; for $Δ$ the constant depends only on $n, α$ . The $n = 1$ calculations below prove the same comparison directly and do not invoke this supplier. (Euclidean spheres and closed balls as subspaces of $R^{n}$ )

[F3]

The chain rule computes the derivatives of the quadratic: for $w (x) = c (R^{2} - ∣ x - x_{0} ∣^{2}) / (2 n)$ one has $D_{i} w (x) = - c (x_{i} - x_{0, i}) / n$ and $D_{i} D_{j} w = - c δ_{i j} / n$ . (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

Verification

technique · direct

1.1F3F1givenalgebra

Derivatives and the equation. By [F3], $Δ u = \sum_{i} D_{i} D_{i} u = - n \cdot c / n = - c$ , so $Δ u \equiv - c$ and $- Δ u \equiv c$ on $B_{R} (x_{0})$ ; moreover $u = 0$ on $\partial B_{R} (x_{0})$ because $∣ x - x_{0} ∣ = R$ there. With $L = Δ$ the datum is $L u = Δ u \equiv - c$ , a constant function on $B_{R} (x_{0})$ .

2.1step 1.1F1algebra

The exact values on the inner ball. Write $r : = ∣ x - x_{0} ∣$ . On $B_{R / 2} (x_{0})$ one has $∣ u (x) ∣ = ∣ c ∣ (R^{2} - r^{2}) / (2 n)$ , maximal at $r = 0$ with value $∣ c ∣ R^{2} / (2 n)$ ; next $∣ D u (x) ∣ = ∣ c ∣ r / n$ , with supremum as $r ↑ R / 2$ equal to $∣ c ∣ R / (2 n)$ ; finally $D^{2} u \equiv - c I_{n} / n$ is constant, so $∣ D^{2} u ∣ \equiv ∣ c ∣ / n$ and the H"older seminorm $[D^{2} u]_{0, α; B_{R / 2} (x_{0})}$ vanishes.

3.1step 2.1F1F2algebra

The scaled norm and the two sides. By the definition in [F1] and step 2.1, $∥ u ∥_{2, α; B_{R / 2} (x_{0})}^{*} = \frac{∣ c ∣ R^{2}}{2 n} + \frac{R}{2} \cdot \frac{∣ c ∣ R}{2 n} + \frac{R^{2}}{4} \cdot \frac{∣ c ∣}{n} + 0 = \frac{∣ c ∣ R^{2}}{n},$ while $∥ u ∥_{\infty; B_{R}} = ∣ c ∣ R^{2} / (2 n)$ (the centre value), $∥ L u ∥_{\infty; B_{R}} = ∣ c ∣$ and $[L u]_{0, α; B_{R}} = 0$ because $L u$ is constant; hence the right-hand side of the estimate of [F2] is $C (\frac{∣ c ∣ R^{2}}{2 n} + ∣ c ∣ R^{2}) = C ∣ c ∣ R^{2} (\frac{1}{2 n} + 1)$ , proportional to the left-hand side with an $R$ -independent factor.

4.1step 1.1step 3.1F1algebra

The dilation identity. Put $v (z) : = u (x_{0} + R z) = c R^{2} (1 - ∣ z ∣^{2}) / (2 n)$ on $B_{1 / 2} (0)$ . The chain rule gives $D^{β} v (z) = R^{∣ β ∣} D^{β} u (x_{0} + R z)$ , and the scaling identity of [F1] yields $∥ v ∥_{2, α; B_{1 / 2} (0)}^{*} = ∥ u ∥_{2, α; B_{R / 2} (x_{0})}^{*}$ ; directly, $∥ v ∥_{2, α; B_{1 / 2} (0)}^{*} = ∣ c ∣ R^{2} (\frac{1}{2 n} + \frac{1}{4 n} + \frac{1}{4 n}) = \frac{∣ c ∣ R^{2}}{n}$ , in agreement with step 3.1.

5.1step 3.1step 4.1F2∎

Conclusion. The quadratic Poisson solution realizes the a priori estimate of [F2] with the same radius homogeneity on both sides: the scaled norm and the scaled data are both of size $∣ c ∣ R^{2}$ , the comparison constant is independent of $R$ , and the dilation identity of the scaled norms holds exactly.

Remarks

The example is the constant-coefficient extremal for the radius bookkeeping: the solution is a parabola, $D^{2} u$ is constant so the top-order H"older seminorm vanishes, and all growth in $R$ comes from the sup terms with their weights $R^{j}$ .
With the sign convention $L u = Δ u$ the right-hand side of the estimate is a bound in terms of $∥ L u ∥_{\infty; B_{R}} = ∣ c ∣$ , exactly as displayed; the value at the centre, $∣ c ∣ R^{2} / (2 n)$ , is the sup over $B_{R}$ , while the sup over the inner ball is the same quantity, since the parabola is maximal at the centre.

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

A non-Dini continuous Poisson source can destroy the continuity of the second derivatives

Statement refuted

The assertion that continuity of a compactly supported source $f$ suffices for the Newtonian potential $N f$ to be of class $C^{2}$ with continuous second derivatives is false; the counterexample and its verification are in the next section.

Facts & Assumptions

Given: ${A C}_{ω}$ , the dimension $n = 2$ , any function $h$ and source $f$ with the properties constructed in the counterexample below, and the sign convention $- Δ Φ = δ_{0}$ with $Φ (x) = - (2 π)^{- 1} \log ∣ x ∣$ of Fundamental solution for the positive operator minus Laplacian.

[A1]

The only choice assumption is Countable Choice ${A C}_{ω}$ , used through the measure, polar and potential interfaces cited below; no full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$f$ is continuous: $∣ f (x) ∣ \leq h (∣ x ∣)$ with $h (r) \to 0$ as $r ↓ 0$ , and $f$ is smooth on ${x \neq 0}$ ; moreover $supp f \subseteq {\overline{B}}_{1 / 3} (0) \subseteq {\overline{B}}_{1 / 2} (0)$ . The Newtonian potential $N f (x) = \int Φ (x - y) f (y) d y$ of the bounded compactly supported $f$ is absolutely finite at every $x$ and locally bounded. (Newtonian potential of compactly supported data, Bounded compact data give an everywhere finite Newtonian potential, Euclidean spheres and closed balls as subspaces of $R^{n}$ )

[F2]

$\partial_{i} Φ (z) = - z_{i} / (2 π ∣ z ∣^{2})$ for $z \neq 0$ , and the mixed kernel is $\partial_{1} \partial_{1} Φ (z) = \cos (2 θ) / (2 π ∣ z ∣^{2})$ with $z = ∣ z ∣ (\cos θ, \sin θ)$ ; polar integration on $R^{2}$ uses $r d r d θ$ . (Fundamental solution for the positive operator minus Laplacian, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F3]

For integrable parameter-dependent functions differentiable in the parameter, differentiation under the integral is valid when the parameter derivatives are measurable and bounded by one integrable majorant throughout a parameter neighbourhood. Repeated differentiation requires this hypothesis at each order. If the resulting derivatives are also pointwise continuous in the parameter under such majorants, dominated convergence makes the parameter integral derivatives continuous. (Differentiation under the integral sign, Dominated convergence)

[F4]

For every $0 < α < 1$ , every compactly supported continuous $g$ with finite $α$ -Hölder seminorm has $N g \in C^{2} (R^{2})$ with second derivatives locally $α$ -Hölder and $- Δ N g = g$ pointwise; in particular this applies to every $g \in C_{c}^{\infty} (R^{2})$ and to every $C_{c}^{0, α}$ function. (Hölder data give a classical Newtonian solution, Local Hölder and scaled C-two-alpha norms on balls)

[F5]

A smooth cutoff equal to one on a compact set and supported in a prescribed larger open set exists (rescalings of a fixed bump), and the mean value theorem bounds an increment by a derivative supremum times the length of the segment. (A smooth bump between concentric Euclidean balls, The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , $C^{k}$ maps and multi-index derivative notation in Euclidean space)

[F6]

If functions converge pointwise under one integrable majorant, their integrals converge (Dominated convergence).

[L0]

The refuted claim: continuity of a compactly supported continuous source $f$ implies that the Newtonian potential $N f$ is $C^{2}$ with continuous second derivatives.

Counterexample

Choose a smooth scalar cutoff $η \in C^{\infty} ([0, \infty); [0, 1])$ with $η = 1$ on $[0, 1 / 4]$ and $η = 0$ on $[1 / 3, \infty)$ . In $R^{2}$ define $h (0) : = 0$ , $h (r) : = η (r) / \log (1 / r)$ for $0 < r < 1 / 3$ , and $h (r) : = 0$ for $r \geq 1 / 3$ . Then $h$ is continuous at $0$ , smooth on $(0, \infty)$ , equals $1 / \log (1 / r)$ for $0 < r \leq 1 / 4$ , and is supported in $[0, 1 / 3]$ . Put $f (x) : = h (∣ x ∣) \frac{x_{1}^{2} - x_{2}^{2}}{∣ x ∣^{2}}$ for $x \neq 0$ and $f (0) : = 0$ . The claim refuted is that such a continuous compactly supported source forces $N f$ to be $C^{2}$ with continuous second derivatives.

Proof technique: direct.

1.1givenF1algebraA1

Continuity, smoothness away from $0$ , and support. Since $(x_{1}^{2} - x_{2}^{2}) / ∣ x ∣^{2}$ is bounded by $1$ and $h \leq 1$ , $∣ f (x) ∣ \leq h (∣ x ∣)$ for $x \neq 0$ and $f (0) = 0$ ; as $h (r) = 1 / \log (1 / r) \to 0$ near $0$ , $f$ is continuous there. For $x \neq 0$ , both $h (∣ x ∣)$ and the angular factor are smooth, so $f$ is smooth there. It vanishes for $∣ x ∣ \geq 1 / 3$ , hence is compactly supported in ${\overline{B}}_{1 / 3} (0)$ and the potential is everywhere absolutely finite by [F1].

2.1step 1.1F1F2F6algebra

$N f \in C^{1}$ and its first derivatives are kernel convolutions. Fix $x \in R^{2}$ , a coordinate $i$ , and $h \neq 0$ , and write $z = x - y$ , $r = ∣ h ∣$ . Then $\frac{N f (x + h e_{i}) - N f (x)}{h} = \int \frac{Φ (z + h e_{i}) - Φ (z)}{h} f (x - z) d z,$ while the candidate derivative is $G_{i} (x) : = \int \partial_{i} Φ (z) f (x - z) d z$ , which is absolutely finite because $∣ \partial_{i} Φ (z) ∣ \leq C ∣ z ∣^{- 1}$ is locally integrable and $f$ is bounded with compact support. On $∣ z ∣ \leq 2 r$ , local integrability of the logarithmic kernel gives $\frac{1}{r} \int_{∣ z ∣ \leq 2 r} (∣ Φ (z + h e_{i}) ∣ + ∣ Φ (z) ∣) ∣ f (x - z) ∣ d z + \int_{∣ z ∣ \leq 2 r} ∣ \partial_{i} Φ (z) f (x - z) ∣ d z \leq C ∥ f ∥_{\infty} r (1 + ∣ \log r ∣) \to 0.$ On $∣ z ∣ > 2 r$ , the segment from $z$ to $z + h e_{i}$ stays at distance at least $∣ z ∣ / 2$ from the origin, so the mean-value estimate for $D^{2} Φ$ gives $∣ \frac{Φ (z + h e_{i}) - Φ (z)}{h} - \partial_{i} Φ (z) ∣ \leq C r ∣ z ∣^{- 2} .$ Since the support of $f (x - \cdot)$ is bounded, the integral of this error is at most $C ∥ f ∥_{\infty} r \int_{2 r}^{R_{x}} s^{- 1} d s = o (1)$ for a fixed finite $R_{x}$ containing that support. Thus $\partial_{i} N f (x) = G_{i} (x)$ . For $x^{'}$ in a fixed compact neighborhood of any $x_{0}$ , choose one ball $B_{R}$ containing all supports of $f (x^{'} - \cdot)$ ; after the change of variables above, $N f (x^{'})$ and $G_{i} (x^{'})$ are integrals on $B_{R}$ dominated respectively by $∥ f ∥_{\infty} ∣ Φ ∣$ and $∥ f ∥_{\infty} ∣ \partial_{i} Φ ∣$ , both integrable there. For any sequence $x_{j} \to x_{0}$ , pointwise continuity of $f$ and [F6] give $N f (x_{j}) \to N f (x_{0})$ and $G_{i} (x_{j}) \to G_{i} (x_{0})$ ; sequential continuity on $R^{2}$ proves continuity of $N f$ and each $G_{i}$ . Hence $N f \in C^{1}$ by the definition of $C^{1}$ . Inserting [F2] gives $\partial_{1} N f (x) = - (2 π)^{- 1} \int (x_{1} - y_{1}) ∣ x - y ∣^{- 2} f (y) d y$ .

2.2step 1.1F2algebra

The truncated Hessian integral. With $z = r (\cos θ, \sin θ)$ and $\cos 2 θ = 2 \cos^{2} θ - 1$ , the angular factor gives $z_{1}^{2} - z_{2}^{2} = r^{2} \cos 2 θ$ , so by [F2] and polar integration, for $0 < ε < 1 / 4$ , $\int_{ε < ∣ z ∣ < 1} \partial_{1} \partial_{1} Φ (- z) f (z) d z = \frac{1}{2 π} \int_{ε}^{1 / 4} \frac{h (r)}{r} d r \int_{0}^{2 π} \cos^{2} 2 θ d θ + O (1) = \frac{1}{2} \int_{ε}^{1 / 4} \frac{h (r)}{r} d r + O (1),$ where $\int_{0}^{2 π} \cos^{2} 2 θ d θ = π$ and the $O (1)$ collects the region $1 / 4 < ∣ z ∣ < 1$ , on which $∣ h ∣ \leq 1$ and $∣ \partial_{1} \partial_{1} Φ ∣ \leq (2 π)^{- 1} ∣ z ∣^{- 2}$ . Since $\int h (r) r^{- 1} d r = - \log \log (1 / r)$ on $(0, 1 / 4)$ , the last expression equals $\frac{1}{2} \log \log (1 / ε) + O (1) \to + \infty$ : the truncated integrals diverge and the principal value $p . v . \int \partial_{1} \partial_{1} Φ (- z) f (z) d z$ does not exist.

2.3step 1.1F3F4F5algebra

Away from the origin second derivatives are finite and continuous. Let $x \neq 0$ and choose, by [F5], a cutoff $χ = 1$ on $\overline{B} (x, ∣ x ∣ / 4)$ with $supp χ \subset B (x, ∣ x ∣ / 2)$ . By step 1.1, $f$ is smooth away from $0$ , so $χ f \in C_{c}^{\infty} \subset C_{c}^{0, 1 / 2}$ ; [F4] gives $N (χ f) \in C^{2}$ near $x$ . For $g : = (1 - χ) f$ , one has $g = 0$ on $B (x, ∣ x ∣ / 4)$ . Thus for $x^{'} \in B (x, ∣ x ∣ / 8)$ and $y \in supp g$ , $∣ x^{'} - y ∣ \geq ∣ x - y ∣ - ∣ x - x^{'} ∣ \geq ∣ x ∣ / 8 > 0$ . The integrand $Φ (x^{'} - y) g (y)$ is smooth in $x^{'}$ there, and it and all its $x^{'}$ -derivatives are dominated by constants times $∣ g (y) ∣$ on the bounded support. Repeated differentiation under the integral sign [F3] shows that $N g$ is smooth on $B (x, ∣ x ∣ / 8)$ . Hence $N f$ is $C^{2}$ near every $x \neq 0$ with finite continuous second derivatives there.

3.1step 2.1step 2.2F2F5algebra

The second derivative at $0$ does not exist. By step 2.1, for $t \neq 0$ , $\frac{\partial_{1} N f (t e_{1}) - \partial_{1} N f (0)}{t} = - \frac{1}{2 π t} \int (\frac{t e_{1} - y}{∣ t e_{1} - y ∣^{2}} + \frac{y}{∣ y ∣^{2}})_{1} f (y) d y .$ Split at $∣ y ∣ = 2 ∣ t ∣$ . On $∣ y ∣ \leq 2 ∣ t ∣$ the integrand is bounded by $C ∥ f ∥_{\infty} (∣ y ∣^{- 1} + ∣ t e_{1} - y ∣^{- 1})$ , whose integral over the ball of radius $2 ∣ t ∣$ is $O (∣ t ∣)$ , so this part contributes $O (1)$ after division by $t$ . On $∣ y ∣ \geq 2 ∣ t ∣$ the second-order Taylor formula along the segment from $- y$ to $t e_{1} - y$ , together with the homogeneity of $\partial_{1} Φ$ (degree $- 1$ ) and the mean value bound [F5] for its second derivatives (degree $- 3$ ), writes the integrand divided by $t$ as $\partial_{1} \partial_{1} Φ (- y) f (y) + O (∣ t ∣ ∣ y ∣^{- 3}) f (y)$ , and $\int_{∣ y ∣ \geq 2 ∣ t ∣} ∣ t ∣ ∣ y ∣^{- 3} ∣ f (y) ∣ d y \leq C ∣ t ∣ \int_{2 ∣ t ∣}^{1 / 2} s^{- 2} d s = O (1)$ . The main term is exactly the integral of step 2.2 with $ε = 2 ∣ t ∣$ , equal to $\frac{1}{2} \log \log (1 / (2 ∣ t ∣)) + O (1) \to + \infty$ as $t \to 0$ . Hence the difference quotients of $\partial_{1} N f$ at $0$ diverge to $+ \infty$ and $\partial_{1} \partial_{1} N f (0)$ does not exist; a fortiori $N f \notin C^{2}$ near the origin, and its second derivatives are not continuous there.

4.1step 1.1step 2.2step 3.1step 2.3given∎

Conclusion. Steps 1.1 and 2.1--3.1 exhibit a continuous compactly supported source $f$ whose Newtonian potential is well defined and $C^{1}$ , is $C^{2}$ away from one point, but fails to be twice differentiable at that point; the truncated Hessian integrals at $0$ diverge like $\frac{1}{2} \log \log (1 / ε)$ . Therefore continuity of the source does not imply continuity of the second derivatives of $N f$ , and the counterexample refutes exactly that overclaim. The Hölder hypothesis of Hölder data give a classical Newtonian solution is used there only through a Dini-type small-scale estimate, and $h (r) = 1 / \log (1 / r)$ has $\int_{0}^{1 / 4} h (r) r^{- 1} d r = + \infty$ , so the failure occurs precisely at the modulus threshold.

Remarks

The source $f$ is continuous and compactly supported but not Hölder continuous at the origin: if $ω_{f} (r) : = \sup_{∣ x - y ∣ \leq r} ∣ f (x) - f (y) ∣$ , then $ω_{f} (r) \geq ∣ f (r e_{1}) - f (0) ∣ = 1 / \log (1 / r)$ for $0 < r \leq 1 / 4$ , so $\int_{0}^{1 / 4} ω_{f} (r) r^{- 1} d r = \infty$ . No exact equality of the modulus with its radial profile is needed. The example therefore isolates the small-scale modulus as the exact input needed by the Newtonian regularity theorem, beyond mere continuity.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The Schauder estimate fails at the H"older endpoint $α = 1$

Statement refuted

The interior Schauder estimate of Interior Schauder estimate for uniformly elliptic equations does not extend to the endpoint $α = 1$ : there is no constant $C < \infty$ such that $[D^{2} N f]_{C^{0, 1} (R^{2})} \leq C ∥ f ∥_{C^{0, 1} (R^{2})}$ for every compactly supported Lipschitz source $f$ and its Newtonian potential $N f$ . The explicit counterexample below uses a Lipschitz source whose angular profile is the degree-one homogeneous mode $r \cos 3 θ$ ; the second derivatives inherit an $r \log r$ term, which tends to zero but is not Lipschitz at the origin. Wang's theory gives the sharp bound with the logarithm $∣ x ∣ ∣ \log ∣ x ∣ ∣$ and the example shows that this logarithm cannot be removed.

Facts & Assumptions

Given: ${A C}_{ω}$ , the cut-off function $χ$ with $χ = 1$ on $[0, 1]$ , $χ (r) = 2 - r$ for $1 < r < 2$ and $χ (r) = 0$ for $r \geq 2$ (with $χ (r) : = 0$ for $r < 0$ ), the source $f (x_{1}, x_{2}) : = χ (r) (x_{1}^{3} - 3 x_{1} x_{2}^{2}) / r^{2}$ for $r > 0$ and $f (0) : = 0$ , the local function $p (r, θ) : = - \frac{1}{6} r^{3} \log r \cos 3 θ$ on $0 < r < 1$ with $p (0) : = 0$ , and its Newtonian potential $N f$ .

[A1]

The only choice principle used is Countable Choice ${A C}_{ω}$ ; no full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The Newtonian potential $N f$ is the convolution integral with the kernel $Φ$ wherever defined (Newtonian potential of compactly supported data); $Φ$ is the kernel candidate for $- Δ$ , with the sign convention fixed in Fundamental solution for the positive operator minus Laplacian. The pointwise equation $- Δ N f = f$ used below is supplied by [F2].

[F2]

For $f \in C_{c}^{0, α} (R^{2})$ , $0 < α < 1$ , the potential $N f$ is $C^{2}$ and satisfies $- Δ N f = f$ pointwise; the local Hölder classes are those of Local Hölder and scaled C-two-alpha norms on balls, and the cancelled representation of the second derivatives is the one of The cancelled representation of the second derivatives of Newtonian potentials. (Hölder data give a classical Newtonian solution)

[F3]

Harmonic functions are real analytic, so on every compact subset of their domain all partial derivatives are bounded and, in particular, the Hessian is Lipschitz. (Harmonic functions are real analytic)

Counterexample

technique · direct

1.1givenalgebraA1

The source is compactly supported and Lipschitz. For $r > 0$ one has $x_{1}^{3} - 3 x_{1} x_{2}^{2} = r^{3} \cos 3 θ$ , so $f (x) = χ (r) r \cos 3 θ$ , a product of the radial function $χ$ with the degree-one homogeneous function $g (x) : = r \cos 3 θ$ . The function $g$ is smooth off the origin, satisfies $∣ g ∣ \leq r$ and, being $1$ -homogeneous, has $∣ \nabla g ∣ \leq C_{0}$ globally, while $χ$ is Lipschitz with support in ${\overset{ˉ}{B}}_{2}$ ; hence $f$ is compactly supported, $\sup ∣ f ∣ \leq 2$ and $[f]_{0, 1} \leq C_{0} ∥ χ ∥_{\infty} + 2 [χ]_{0, 1} < \infty$ , that is $f \in C_{c}^{0, 1} (R^{2})$ with finite $C^{0, 1}$ norm.

1.2givenalgebra

The explicit local solution. Put $P (x) : = x_{1}^{3} - 3 x_{1} x_{2}^{2}$ , so that $P = r^{3} \cos 3 θ$ and $p = - \frac{1}{6} P \log r$ for $0 < r < 1$ . The polynomial $P$ is harmonic: $\partial_{11} P = 6 x_{1}$ , $\partial_{22} P = - 6 x_{1}$ , hence $Δ P = 0$ ; moreover $\nabla P \cdot \nabla \log r = P_{i} x_{i} / r^{2}$ with $P_{i} x_{i} = 3 (x_{1}^{2} - x_{2}^{2}) x_{1} + (- 6 x_{1} x_{2}) x_{2} = 3 x_{1} (x_{1}^{2} - 3 x_{2}^{2}) = 3 r^{3} \cos 3 θ$ , so $\nabla P \cdot \nabla \log r = 3 r \cos 3 θ$ ; and $Δ \log r = 0$ for $r > 0$ . Therefore, for $0 < r < 1$ , $Δ p = - \frac{1}{6} (Δ P \log r + 2 \nabla P \cdot \nabla \log r + P Δ \log r) = - \frac{1}{6} \cdot 2 \cdot 3 r \cos 3 θ = - r \cos 3 θ = - f (x),$ because $χ (r) = 1$ there. At the origin $p$ is $C^{2}$ with $D^{2} p (0) = 0$ : indeed $p = O (r^{3} ∣ \log r ∣)$ , $\partial_{i} p = O (r^{2} ∣ \log r ∣)$ and $\partial_{i} \partial_{j} p = O (r ∣ \log r ∣)$ as $r ↓ 0$ , as the three terms of the product rule show, so $Δ p (0) = 0 = f (0)$ and the identity $- Δ p = f$ holds pointwise on all of the unit disc.

2.1step 1.2F1F2F3algebra

The potential differs from $p$ by a harmonic function. The source $f$ is Lipschitz, hence belongs to $C_{c}^{0, α} (R^{2})$ for every $0 < α < 1$ , so by [F2] the potential $N f$ is $C^{2}$ with $- Δ N f = f$ pointwise. Step 1.2 gives $- Δ p = f$ pointwise on the unit disc, so $Δ (N f - p) = 0$ there; by [F3] the function $H : = N f - p$ is real analytic on the unit disc and its Hessian is Lipschitz on $B_{1 / 2} (0)$ , say $∣ D^{2} H (x) - D^{2} H (y) ∣ \leq M ∣ x - y ∣$ for $x, y \in B_{1 / 2} (0)$ .

3.1step 1.1step 1.2step 2.1algebra

Failure of the Lipschitz bound. On the positive $x_{1}$ -axis $p (x_{1}, 0) = - \frac{1}{6} x_{1}^{3} \log x_{1}$ for $0 < x_{1} < 1$ , so $\partial_{11} p (x_{1}, 0) = - \frac{1}{6} (6 x_{1} \log x_{1} + 5 x_{1}) = - x_{1} \log x_{1} - \frac{5}{6} x_{1}, \partial_{11} p (0, 0) = 0.$ By step 2.1, $\partial_{11} N f (x_{1}, 0) = - x_{1} \log x_{1} - \frac{5}{6} x_{1} + \partial_{11} H (x_{1}, 0)$ and the last term deviates from its value at $0$ by at most $M x_{1}$ . Hence for $0 < x_{1} < \frac{1}{2}$ , $\frac{∣ \partial_{11} N f (x_{1}, 0) - \partial_{11} N f (0, 0) ∣}{x_{1}} \geq ∣ \log x_{1} ∣ - \frac{5}{6} - M ⟶ + \infty (x_{1} ↓ 0) .$ Therefore $[D^{2} N f]_{0, 1} = \sup_{x \neq y} ∣ D^{2} N f (x) - D^{2} N f (y) ∣ / ∣ x - y ∣ \geq {lim sup}_{x_{1} ↓ 0} (\dots) = + \infty$ , while $∥ f ∥_{C^{0, 1}} = \sup ∣ f ∣ + [f]_{0, 1} < \infty$ by step 1.1.

4.1step 1.1step 3.1given∎

Conclusion. The compactly supported Lipschitz source $f$ of step 1.1 has finite $C^{0, 1}$ norm, but its Newtonian potential has $[D^{2} N f]_{0, 1} = + \infty$ by step 3.1; hence no finite constant $C$ can satisfy $[D^{2} N f]_{C^{0, 1} (R^{2})} \leq C ∥ f ∥_{C^{0, 1} (R^{2})}$ , and the endpoint $α = 1$ version of the Schauder estimate is false. The example is consistent with the true sharp result: $D^{2} N f$ is bounded and has the logarithmic modulus $∣ x ∣ ∣ \log ∣ x ∣ ∣$ , so the failure is exactly the loss of one logarithm, not a loss of boundedness.

Remarks

The computation is the standard sharpness construction: the degree-one homogeneous forcing $r \cos 3 θ$ produces a degree-three logarithmic potential, and the positive axis is where the $\log$ term is visible. The angular factor is immaterial; the angular mode $\cos 3 θ$ is resonant with the degree-three radial ansatz. A degree-one spherical harmonic instead gives linear forcing and does not produce this logarithmic obstruction.
The example refutes the endpoint case of the interior estimate for the Laplacian; it does not contradict the strict-range estimate for $0 < α < 1$ , which is proved for compactly supported H"older data and has no uniform Lipschitz-endpoint constant. Wang equation (1.4) bounds the Hessian increment by $C_{n} d (\sup ∣ u ∣ + ∥ f ∥_{C^{0, 1}} ∣ \log d ∣)$ , with $d = ∣ x - y ∣$ .

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The Riesz-transform formula for second derivatives of the Laplacian

Example

Assume Countable Choice. Let $n \geq 1$ and $u \in S (R^{n})$ . Then, as tempered distributions (equivalently, as $L^{2}$ classes almost everywhere), $\partial_{i} \partial_{j} u = - R_{i} R_{j} (Δ u) = R_{i} R_{j} (- Δ u) (i, j \in {1, \dots, n}),$ where $R_{1}, \dots, R_{n}$ are the Riesz transforms of Riesz transforms on Euclidean space; the symbol of $R_{i} R_{j}$ is $- ξ_{i} ξ_{j} / ∣ ξ ∣^{2}$ off the origin, and combining the $L^{p}$ bounds of The Riesz transforms are bounded on Lp gives $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} \leq C_{n, p}^{2} ∥ Δ u ∥_{L^{p}} (1 < p < \infty) .$ For $n = 1$ the identity reads $u^{''} = - R_{1} R_{1} u^{''}$ and is consistent because $R_{1}^{2} = - i d$ makes both sides equal $u^{''}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , $n \geq 1$ , $1 < p < \infty$ , a Schwartz function $u \in S (R^{n})$ , and the negative-sign $2 π$ -normalized Fourier convention.

[A1]

The only choice assumption is Countable Choice ${A C}_{ω}$ ; it enters through the Plancherel and Riesz-transform interfaces. No full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The Riesz transforms are $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ with $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ and $m_{j} (0) = 0$ ; $∥ R_{j} ∥_{L^{2} \to L^{2}} \leq 1$ and $\sum_{j} R_{j}^{2} = - i d$ on $L^{2}$ . For $1 < p < \infty$ each $R_{j}$ extends boundedly to $L^{p} (R^{n}; C)$ with norm at most $C_{n, p}$ . (Riesz transforms on Euclidean space, Riesz transforms are L2 contractions and square to minus the identity in sum, The Riesz transforms are bounded on Lp, Exact L2 Fourier multiplier norm)

[F2]

$F (\partial^{α} u) = (2 π i ξ)^{α} F u$ on $S^{'} (R^{n})$ , and $F_{2}$ is a linear isometry that is injective on $L^{2}$ ; two tempered distributions with the same Fourier transform are equal. (Fourier differentiation and multiplication identities on tempered distributions, Plancherel theorem)

Verification

technique · direct

1.1F1F2algebraA1

Fourier multipliers. For $u \in S (R^{n})$ and any $i, j$ , [F2] gives $F (\partial_{i} \partial_{j} u) = (2 π i ξ_{i}) (2 π i ξ_{j}) \hat{u} = - 4 π^{2} ξ_{i} ξ_{j} \hat{u}$ and $F (Δ u) = - 4 π^{2} ∣ ξ ∣^{2} \hat{u}$ in $S^{'}$ ; since $u$ and its derivatives are Schwartz functions, these are also the $L^{2}$ Fourier transforms $F_{2}$ of the corresponding classes. Off the origin $m_{i} (ξ) m_{j} (ξ) = (- i ξ_{i} / ∣ ξ ∣) (- i ξ_{j} / ∣ ξ ∣) = - ξ_{i} ξ_{j} / ∣ ξ ∣^{2}$ , so $m_{i} m_{j} \cdot (- 4 π^{2} ∣ ξ ∣^{2}) = 4 π^{2} ξ_{i} ξ_{j}$ and hence $F_{2} (R_{i} R_{j} (- Δ u)) = F_{2} (\partial_{i} \partial_{j} u)$ almost everywhere: indeed $R_{i} R_{j} (- Δ u) = F_{2}^{- 1} (m_{i} m_{j} F_{2} (- Δ u))$ and $F_{2} (- Δ u) = 4 π^{2} ∣ ξ ∣^{2} \hat{u}$ , while the value of $m_{i} m_{j}$ at the single point $ξ = 0$ is immaterial. Injectivity of $F_{2}$ [F2] gives the $L^{2}$ identity $\partial_{i} \partial_{j} u = R_{i} R_{j} (- Δ u)$ , hence also the distributional identity and the sign rearrangement $R_{i} R_{j} (- Δ u) = - R_{i} R_{j} (Δ u)$ .

2.1step 1.1F1algebra

Symbol and $L^{p}$ bound. The multiplier of $R_{i} R_{j}$ is $m_{i} m_{j} = - ξ_{i} ξ_{j} / ∣ ξ ∣^{2}$ off the origin, so its absolute value is at most $1$ ; applying [F1] twice and using the identity of step 1.1, $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} = ∥ R_{i} R_{j} (- Δ u) ∥_{L^{p}} \leq C_{n, p}^{2} ∥ Δ u ∥_{L^{p}}$ for $1 < p < \infty$ , the norms being those of the $L^{p}$ classes of the Schwartz functions involved.

2.2step 1.1F1algebra

The one-dimensional case. For $n = 1$ one has $m_{1} (ξ) = - i ξ / ∣ ξ ∣ = - i sign (ξ)$ off the origin, so $m_{1}^{2} = - 1$ and therefore $R_{1}^{2} = - i d$ on $L^{2}$ by [F1]; the identity of step 1.1 then reads $u^{''} = R_{1} R_{1} (- u^{''}) = - R_{1} R_{1} u^{''}$ , whose right-hand side equals $u^{''}$ , so the two sides agree.

3.1step 1.1step 2.1step 2.2given∎

Conclusion. For every $n \geq 1$ and $u \in S (R^{n})$ the second derivatives are the composition of the second-order Riesz multiplier with $- Δ u$ ; the strict range $1 < p < \infty$ is inherited from the Riesz-transform $L^{p}$ theorem, and the sign convention is the negative-sign $2 π$ -normalized Fourier transform used throughout. No endpoint $p = 1$ or $p = \infty$ bound is asserted.

Remarks

The formula identifies the Hessian of $u$ with a bounded combination of Riesz transforms of the Laplacian, which is the multiplier version of the Calderón–Zygmund representation of second derivatives; it is the whole-space model estimate behind the interior $W^{2, p}$ regularity on this page.

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Boundary $W^{2, p}$ regularity needs more than Lipschitz boundary

Statement refuted

A weak-solution regularity assertion that replaces the $C^{1, 1}$ boundary hypothesis by mere Lipschitz regularity is false; this does not refute the a priori estimate Global $W^{2, p}$ Dirichlet estimate on a $C^{1, 1}$ domain, whose hypothesis already requires $u \in W^{2, p}$ . The reentrant sector below is a bounded Lipschitz domain, but not a $C^{1, 1}$ domain at its vertex. Let $Ω_{ω} = {(r, θ) : 0 < r < 1, 0 < θ < ω}$ with $ω \in (π, 2 π)$ , put $γ = π / ω \in (1 / 2, 1)$ , and define $U = r^{γ} \sin (γ θ)$ . Choose a smooth radial cutoff $ζ$ supported in $B_{1 / 2}$ and equal to $1$ near $0$ , and let $v = ζ U$ . Choose $r_{0} > 0$ so that $ζ = 1$ for $r < r_{0}$ . Then $v \in H_{0}^{1} (Ω_{ω})$ , $f : = - Δ v \in C^{\infty} (Ω_{ω}) \cap L^{\infty} (Ω_{ω}) \subset L^{p} (Ω_{ω})$ for every finite $p$ , and $- Δ v = f$ weakly. However, $∣ D^{2} v ∣ ≍ r^{γ - 2}$ near $0$ , so $\int_{0}^{r_{0}} r^{p (γ - 2) + 1} d r = \infty ⟺ p \geq \frac{2}{2 - γ} = \frac{2 ω}{2 ω - π} .$ Hence this weak solution is not in $W^{2, p}$ for those exponents (in particular not in $H^{2}$ ). The reentrant corner shows why weak boundary regularity requires more than a Lipschitz chart.

Facts & Assumptions

Given: ${A C}_{ω}$ , $n = 2$ , $ω \in (π, 2 π)$ , $γ = π / ω$ , the sector $Ω_{ω} = {(r, θ) : 0 < r < 1, 0 < θ < ω}$ (with $x = r \cos θ$ , $y = r \sin θ$ ), the harmonic profile $U (r, θ) = r^{γ} \sin (γ θ)$ , a radial cutoff $ζ \in C_{c}^{\infty} (R^{2})$ with $ζ = 1$ on $B_{r_{0}}$ , $0 \leq ζ \leq 1$ and $supp ζ \subseteq B_{1 / 2}$ , and $v = ζ U$ .

[A1]

The only choice principle used is Countable Choice ${A C}_{ω}$ ; no full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Weak solutions of $- Δ u = f$ with zero boundary values are the classes $u \in H_{0}^{1} (Ω_{ω})$ with $\int_{Ω_{ω}} \nabla u \cdot \nabla φ = \int_{Ω_{ω}} f φ$ for every $φ \in H_{0}^{1} (Ω_{ω})$ ; by density it suffices to test against $φ \in C_{c}^{\infty} (Ω_{ω})$ . (Weak Dirichlet solutions for a divergence-form operator, The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field)

[F2]

In polar coordinates on the open sector, $Δ w = w_{r r} + \frac{1}{r} w_{r} + \frac{1}{r^{2}} w_{θ θ}$ for $w \in C^{2}$ ; the Lebesgue integral of a radial function is $\int_{B_{r_{0}} \cap Ω_{ω}} g (r) d x = ω \int_{0}^{r_{0}} g (r) r d r$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F3]

For every $0 < ρ < R$ there is a radial cutoff $ζ \in C_{c}^{\infty} (R^{2})$ with $ζ = 1$ on $B_{ρ}$ , $0 \leq ζ \leq 1$ and $supp ζ \subseteq B_{R}$ . (A smooth bump between concentric Euclidean balls)

[F4]

A class lies in $W^{2, p}$ precisely when it and all its weak derivatives through order two lie in $L^{p}$ . In particular, failure of $L^{p}$ integrability of a second weak derivative excludes $W^{2, p}$ membership; $\int_{0}^{r_{0}} r^{s} d r$ converges if and only if $s > - 1$ . (Integer-order Sobolev spaces and their norms)

Counterexample

technique · direct

1.1F2givenalgebraA1

The profile is harmonic and vanishes on the two sides. For $U (r, θ) = r^{γ} \sin (γ θ)$ one has, by [F2], $Δ U = [γ (γ - 1) r^{γ - 2} + γ r^{γ - 2} - γ^{2} r^{γ - 2}] \sin (γ θ) = 0,$ so $U$ is harmonic on the sector (in particular $U \in C^{\infty} (\overline{Ω_{ω}} ∖ {0})$ ). Moreover $U (r, 0) = 0$ and $U (r, ω) = r^{γ} \sin (π) = 0$ , so $U$ vanishes on the two radial sides of $Ω_{ω}$ ; at the reentrant vertex the sector has interior angle $ω \in (π, 2 π)$ , so $Ω_{ω}$ is a bounded Lipschitz domain that is not $C^{1, 1}$ there.

1.2F1givenalgebraF3

The localized profile is in $H_{0}^{1}$ . Since $∣ U ∣ \leq r^{γ}$ and $∣ \nabla U ∣ = γ r^{γ - 1}$ (the gradient of the harmonic profile has absolute value $γ r^{γ - 1}$ because the angular factor contributes a unit vector in polar coordinates), the integrals $\int_{r < 1 / 2} ∣ U ∣^{2} d x$ and $\int_{r < 1 / 2} ∣ \nabla U ∣^{2} d x$ converge; hence $v = ζ U \in H^{1} (Ω_{ω})$ with support in $\overline{B_{1 / 2}}$ . To approximate $v$ in $H^{1}$ by $C_{c}^{\infty} (Ω_{ω})$ functions: first truncate radially, $v_{ε} : = χ (r / ε) v$ with $χ \in C^{\infty}$ equal to $0$ near $0$ and to $1$ for every $t \geq 2$ . This transition need not be compactly supported: $v_{ε}$ remains compactly supported because $v$ is supported in $\overline{B_{1 / 2}}$ . The estimate $∥ v - v_{ε} ∥_{H^{1}}^{2} = O (ε^{2 γ})$ follows from $∣ v ∣^{2} + ∣ \nabla v ∣^{2} ≲ r^{2 γ - 2}$ near the vertex. Next cut off in the angular variable with a smooth $η_{δ} (θ)$ vanishing for $θ < δ$ and for $θ > ω - δ$ and equal to $1$ for $2 δ < θ < ω - 2 δ$ . For each fixed $ε$ , the squared $H^{1}$ error is $O (δ)$ , hence the $H^{1}$ norm error is $O (δ^{1 / 2})$ : near either side $U = O (r^{γ} dist (θ, {0, ω}))$ , the angular cutoff derivative is $O (δ^{- 1})$ on strips of angular width $O (δ)$ , and the resulting radial weight $r^{2 γ - 1}$ is integrable. The resulting functions are supported in a compact subset of the open sector and can be mollified there, so $v \in H_{0}^{1} (Ω_{ω})$ by definition of the closure.

2.1step 1.2F1givenalgebra

The forcing is smooth and bounded, and the weak equation holds. Since $U$ is harmonic, $Δ v = Δ (ζ U) = (Δ ζ) U + 2 \nabla ζ \cdot \nabla U$ on the sector; the right-hand side is supported in the annulus ${r_{0} \leq r \leq 1 / 2}$ where $U$ and its gradient are smooth up to the two radial sides for $r \geq r_{0} > 0$ , so $f : = - Δ v \in C^{\infty} (Ω_{ω}) \cap L^{\infty} (Ω_{ω})$ and hence $f \in L^{p} (Ω_{ω})$ for every finite $p$ . For $φ \in C_{c}^{\infty} (Ω_{ω})$ integration by parts on the compactly contained support gives $\int \nabla v \cdot \nabla φ = \int (- Δ v) φ = \int f φ$ ; both sides are continuous in $φ$ in the $H^{1}$ norm, so the identity holds for every $φ \in H_{0}^{1} (Ω_{ω})$ by [F1]: $v$ is a weak solution of $- Δ v = f$ with zero boundary values.

3.1step 1.1step 2.1F2F4algebra

The second derivatives diverge exactly above the threshold. On ${r < r_{0}}$ one has $v = U$ , a function homogeneous of degree $γ$ ; write $U = Im z^{γ}$ on the sector branch. Direct differentiation gives $U_{11} = γ (γ - 1) r^{γ - 2} \sin ((γ - 2) θ)$ , $U_{12} = γ (γ - 1) r^{γ - 2} \cos ((γ - 2) θ)$ and $U_{22} = - U_{11}$ . Thus the Frobenius Hessian norm is $\sqrt{2} γ (1 - γ) r^{γ - 2}$ , so there are constants $0 < c_{1} \leq c_{2} < \infty$ with $c_{1} r^{γ - 2} \leq ∣ D^{2} v ∣ \leq c_{2} r^{γ - 2}$ on $Ω_{ω} \cap {0 < r < r_{0}}$ (by the displayed nonvanishing norm, since $0 < γ < 1$ ). By [F2], $\int_{Ω_{ω} \cap B_{r_{0}}} ∣ D^{2} v ∣^{p} d x ≍ \int_{0}^{r_{0}} r^{p (γ - 2) + 1} d r,$ which by [F4] diverges exactly when $p (γ - 2) + 1 \leq - 1$ , that is $p \geq \frac{2}{2 - γ}$ . Using $γ = π / ω$ gives $\frac{2}{2 - γ} = \frac{2 ω}{2 ω - π}$ , so the weak solution $v$ is not in $W^{2, p} (Ω_{ω})$ for those exponents; taking $p = 2$ (which is allowed because $γ < 1$ ) shows in particular that $v \notin H^{2} (Ω_{ω})$ .

4.1step 2.1step 3.1step 1.1given∎

Conclusion. On the bounded Lipschitz reentrant domain $Ω_{ω}$ there is a weak solution $v \in H_{0}^{1} (Ω_{ω})$ of $- Δ v = f$ with $f \in C^{\infty} \cap L^{\infty}$ , which fails to lie in $W^{2, p} (Ω_{ω})$ for every $p \geq 2 ω / (2 ω - π)$ . Thus smooth data and a Lipschitz boundary alone do not guarantee weak-solution $W^{2, p}$ regularity. This example is not a counterexample to the a priori estimate Global $W^{2, p}$ Dirichlet estimate on a $C^{1, 1}$ domain, whose domain already assumes $u \in W^{2, p}$ ; it makes no claim that the estimate’s $C^{1, 1}$ boundary hypothesis is necessary.

Remarks

The mechanism is the corner exponent $γ = π / ω$ : the harmonic profile grows like $r^{γ}$ , its first derivatives like $r^{γ - 1}$ (square-integrable already for $γ > 0$ , since the radial gradient integral is $\int r^{2 γ - 1} d r$ ; the zero-boundary closure was proved in step 1.2), and its second derivatives like $r^{γ - 2}$ , which is not $p$ -integrable for large $p$ because the radial weight in two dimensions is $r^{p (γ - 2) + 1}$ (equal to $r^{2 γ - 3}$ when $p = 2$ ).
The failure is purely at the vertex, not at the sides: the two radial sides are straight, and on each of them the localized solution is smooth for $r \geq r_{0}$ . This isolates the reentrant corner as the obstruction, in contrast to convex corners, where $γ > 1$ improves the integrability threshold, but the Hessian is still unbounded when $1 < γ < 2$ ; it is bounded when $γ \geq 2$ .

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The method of continuity on a constant-coefficient one-dimensional path

Example

Assume Countable Choice and fix $0 < α < 1$ . Let $Ω = (0, π)$ , $X : = {u \in C^{2, α} ([0, π]) : u (0) = u (π) = 0}$ , $Y : = C^{0, α} ([0, π])$ and, for $c > 0$ and $t \in [0, 1]$ , $L_{t} u : = - u^{''} - t c u$ . Then every $L_{t}$ is a bounded operator $X \to Y$ , $L_{0} = - d^{2} / d x^{2}$ is bijective, and the bijectivity set is $I = {t \in [0, 1] : t c \notin {k^{2} : k \geq 1}}$ : for $t c < 1$ the eigenfunction expansion $u (x) = \sum_{k \geq 1} \frac{f_{k}}{k^{2} - t c} \sin (k x), f_{k} = \frac{2}{π} \int_{0}^{π} f (x) \sin (k x) d x,$ converges absolutely and uniformly together with its first derivative, defines an element of $X$ with $L_{t} u = f$ and obeys the uniform bound $∥ u ∥_{C^{2, α}} \leq C (1 - t c)^{- 1} ∥ f ∥_{C^{0, α}}$ , while at $t c = k_{0}^{2}$ the kernel is spanned by $\sin (k_{0} x)$ and the range is the proper closed subspace ${f : \int_{0}^{π} f (x) \sin (k_{0} x) d x = 0}$ . In this model one computes directly that $I$ is open in $[0, 1]$ , that $I$ is relatively closed on every subinterval on which all the operators are injective, and that the uniform estimate fails on every interval that meets the spectrum.

Facts & Assumptions

Given: Countable Choice, $0 < α < 1$ , $c > 0$ , $t \in [0, 1]$ , the spaces $X = {u \in C^{2, α} ([0, π]) : u (0) = u (π) = 0}$ and $Y = C^{0, α} ([0, π])$ , and $L_{t} u = - u^{''} - t c u$ .

[A1]

The only choice assumption is Countable Choice ${A C}_{ω}$ ; all series and subsequences below are countable and no further selection is made. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The norms on $X$ and $Y$ are the usual ones: $∥ u ∥_{C^{2, α}} = \sup ∣ u ∣ + \sup ∣ u^{'} ∣ + \sup ∣ u^{''} ∣ + [u^{''}]_{0, α}$ and $∥ f ∥_{C^{0, α}} = \sup ∣ f ∣ + [f]_{0, α}$ with $[g]_{0, α} = \sup_{x \neq y} ∣ g (x) - g (y) ∣ / ∣ x - y ∣^{α}$ . (Hölder spaces $C^{k, α}$ , closure and interior scaled norms, and $C^{k, α}$ domains)

[F2]

The functions $e_{k} (x) : = \sin (k x)$ , $k \geq 1$ , satisfy $e_{k} (0) = e_{k} (π) = 0$ , $- e_{k}^{''} = k^{2} e_{k}$ , and $\int_{0}^{π} e_{j} (x) e_{k} (x) d x = \frac{π}{2} δ_{j k}$ ; these are the classical eigenpairs of $- d^{2} / d x^{2}$ with Dirichlet conditions on $(0, π)$ . For the odd $2 π$ -periodic extension $F$ , translation by $h$ gives $∥ F (\cdot + h) - F ∥_{L^{1} (- π, π)} \leq C ([f]_{α} h^{α} + ∥ f ∥_{\infty} h)$ : away from endpoint jumps use Hölder continuity, and the jump-crossing strips have length $O (h)$ . With $h = π / k$ , the exponential Fourier coefficient identity $∣ e^{i k h} - 1 ∣ ∣ \hat{F} (k) ∣ \leq (2 π)^{- 1} ∥ F (\cdot + h) - F ∥_{1}$ gives $∣ \hat{F} (k) ∣ \leq C_{α} ∥ f ∥_{C^{0, α}} k^{- α}$ . The sine coefficients $f_{k} = \frac{2}{π} \int_{0}^{π} f (x) \sin (k x) d x$ of an $f \in C^{0, α} ([0, π])$ satisfy $∣ f_{k} ∣ \leq C_{α} ∥ f ∥_{C^{0, α}} k^{- α}$ , and Dini pointwise convergence criterion for Fourier series, after rescaling to period one, gives $\sum_{k \geq 1} f_{k} \sin (k x) = f (x)$ for every $x \in (0, π)$ (the local Dini integral is bounded by $C [f]_{α} \int_{0}^{δ} s^{α - 1} d s$ ). The $L^{2}$ convergence follows separately from Fourier series converge in mean square applied to the odd extension. (Hölder spaces $C^{k, α}$ , closure and interior scaled norms, and $C^{k, α}$ domains)

[F3]

If $g \in C^{0} ([0, π])$ then $S g (x) : = - \int_{0}^{x} (x - s) g (s) d s + \frac{x}{π} \int_{0}^{π} (π - s) g (s) d s$ lies in $C^{2} ([0, π])$ , vanishes at $0$ and $π$ , satisfies $(S g)^{''} = - g$ , and obeys $∥ S g ∥_{C^{2}} \leq C_{0} ∥ g ∥_{C^{0}}$ with $[S g^{''}]_{0, α} = [g]_{0, α}$ for $g \in Y$ ; this is the explicit Dirichlet solution of the one-dimensional Poisson problem, obtained by differentiating twice under the integral sign.

[F4]

Uniform derivative limits: if $u_{N} : [0, π] \to R$ is $C^{1}$ for every $N$ , $u_{N}$ converges at one point, and $u_{N}^{'} \to v$ uniformly, then $u_{N} \to u$ uniformly for a differentiable $u$ with $u^{'} = v$ ; applied twice it gives: if $u_{N} \to u$ uniformly, $u_{N}^{'} \to u^{'}$ uniformly and $u_{N}^{''} \to u^{''}$ uniformly, then $u \in C^{2} ([0, π])$ with those derivatives. (If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit)

[F5]

The abstract method-of-continuity theorem assumes a uniform a priori estimate and Countable Choice (The method of continuity for a uniformly estimated affine family of bounded operators).

[F6]

The global Schauder solvability theorem is the PDE-level version of the continuity argument (Global Schauder estimate and classical Dirichlet solvability by the continuity method).

[F7]

Classical derivatives agree with distributional derivatives under the assumed Countable Choice; distributional differentiation is continuous in the distribution topology (Distributional derivative, Distributional differentiation is continuous and commutes).

[F8]

On the bounded interval, $L^{2}$ convergence implies local $L^{1}$ convergence by Cauchy--Schwarz, and locally $L^{1}$ convergence gives convergence of the associated regular distributions (Locally integrable functions embed in distributions).

[F9]

A distribution on the connected interval $(0, π)$ whose derivative vanishes is a constant regular distribution; this result uses Countable Choice for the regular-distribution convention (A distribution with zero derivatives on a connected open set is constant).

Verification

technique · direct

1.1F1F3givenalgebraA1

The operators and the base point. For $u \in X$ one has $∥ L_{t} u ∥_{C^{0, α}} \leq ∥ u^{''} ∥_{C^{0, α}} + t c ∥ u ∥_{C^{0, α}} \leq C (α, π) (1 + c) ∥ u ∥_{C^{2, α}}$ , so $L_{t}$ maps $X$ boundedly into $Y$ for every $t$ . For $t = 0$ , $L_{0} = - d^{2} / d x^{2}$ : if $- u^{''} = 0$ with $u (0) = u (π) = 0$ then $u$ is affine and vanishes at both endpoints, so $u = 0$ (injectivity); and for every $f \in Y$ the explicit function $S f$ of [F3] satisfies $- S f^{''} = f$ , vanishes at the endpoints, and obeys $∥ S f ∥_{C^{2, α}} \leq C ∥ f ∥_{C^{0, α}}$ , so $L_{0} S f = f$ (surjectivity). Hence $L_{0}$ is bijective.

1.2F2givenalgebra

The eigenvalue picture. By [F2], $L_{t} e_{k} = (k^{2} - t c) e_{k}$ . For any $u \in X$ , two integrations by parts, using $u = e_{k} = 0$ at both endpoints, give $\int_{0}^{π} (L_{t} u) e_{k} = (k^{2} - t c) \int_{0}^{π} u e_{k}$ . If $L_{t} u = 0$ , all sine coefficients of $u$ vanish when $t c$ is not a square; if $t c = k_{0}^{2}$ , all except the $k_{0}$ th vanish. Since $u \in C^{0, α}$ , the Fourier identity in [F2] then gives $u = 0$ in the first case and $u \in span {e_{k_{0}}}$ in the second. Thus $L_{t}$ is injective exactly off the displayed spectrum, and its kernel at a spectral parameter is exactly $span {e_{k_{0}}}$ .

1.3F2F4givenalgebra

The Fourier solution away from resonance, including the coercive range. Fix $f \in Y$ . If $t c = k_{0}^{2}$ , assume $f_{k_{0}} = 0$ and set $c_{k_{0}} = 0$ ; for every $k \in J : = {k \geq 1 : k^{2} \neq t c}$ set $c_{k} = f_{k} / (k^{2} - t c)$ . In the non-resonant case this defines every $c_{k}$ . In either case $m : = \inf_{k \in J} ∣ k^{2} - t c ∣ / k^{2} > 0$ , because the ratios tend to $1$ and none of the finitely many remaining ratios is zero. Put $u_{N} = \sum_{k \leq N} c_{k} \sin (k x)$ . The coefficient bound in [F2] gives $∣ c_{k} ∣ \leq C_{α} m^{- 1} ∥ f ∥_{C^{0, α}} k^{- 2 - α}$ and $k ∣ c_{k} ∣ \leq C_{α} m^{- 1} ∥ f ∥_{C^{0, α}} k^{- 1 - α}$ . Thus only the series for $u_{N}$ and $u_{N}^{'}$ are asserted to converge absolutely and uniformly; [F4] gives a limit $u \in C^{1} ([0, π])$ with zero endpoint values. For $t c < 1$ one has $m \geq 1 - t c$ , giving the stated coercive-range bound. To see $u \in C^{0, α}$ , write $d = ∣ x - y ∣$ . For $0 < d < 1$ , split $\sum_{k} ∣ c_{k} ∣ \min (2, k d)$ at $k \leq d^{- 1}$ : the low-frequency part is at most $C m^{- 1} ∥ f ∥ d \sum_{k \leq d^{- 1}} k^{- 1 - α} \leq C m^{- 1} ∥ f ∥ d$ , and the high-frequency part is at most $C m^{- 1} ∥ f ∥ \sum_{k > d^{- 1}} k^{- 2 - α} \leq C m^{- 1} ∥ f ∥ d^{1 + α}$ . For $1 \leq d \leq π$ , the supremum bound gives the same $C m^{- 1} ∥ f ∥ d^{α}$ control. Hence $[u]_{0, α} \leq C m^{- 1} ∥ f ∥_{C^{0, α}}$ .

2.1step 1.3F2F7F8F9givenalgebra

Identify the equation and upgrade regularity. Put $f_{N} = \sum_{k \leq N} f_{k} \sin (k x)$ . At a resonance the omitted coefficient is zero by hypothesis, so for every sufficiently large $N$ the partial-sum identity is still $u_{N}^{''} = - f_{N} - t c u_{N}$ . By [F2], $f_{N} \to f$ in $L^{2} (0, π)$ , while $u_{N} \to u$ uniformly; hence $u_{N}^{''} \to g : = - f - t c u$ in $L^{2}$ . Since also $u_{N} \to u$ in $L^{2}$ , continuity of distributional differentiation and the regular-function embedding in [F7--F8] show $u^{''} = g$ distributionally on $(0, π)$ . The function $g$ is continuous. Set $G (x) = \int_{0}^{x} g (s) d s$ ; then the distributional derivative of the continuous function $u^{'} - G$ is zero. By [F9], $u^{'} - G$ is a constant distribution, hence equals that constant pointwise; therefore $u \in C^{2} ([0, π])$ (with one-sided endpoint derivatives) and $u^{''} = g$ . Since $u \in C^{0, α}$ by step 1.3 and $f \in C^{0, α}$ , $g = - f - t c u \in C^{0, α}$ , so $u \in X$ and $L_{t} u = f$ .

3.1step 1.2step 1.3step 2.1givenalgebra

The range at a spectral parameter. If $t c = k_{0}^{2}$ , integration by parts as in step 1.2 gives $\int_{0}^{π} (L_{t} u) e_{k_{0}} = 0$ for every $u \in X$ , so the range lies in the proper closed hyperplane ${f \in Y : f_{k_{0}} = 0}$ . Conversely, for any $f$ in that hyperplane, steps 1.3 and 2.1 construct $u \in X$ with $L_{t} u = f$ ; thus this hyperplane is exactly the range. Moreover, the inverse norm of $L_{t}$ on its bijective parameters blows up near $t_{0} : = k_{0}^{2} / c$ : for $t \neq t_{0}$ , testing on $e_{k_{0}}$ gives $∥ L_{t} e_{k_{0}} ∥_{C^{0, α}} = ∣ k_{0}^{2} - t c ∣ ∥ e_{k_{0}} ∥_{C^{0, α}}$ and hence $∥ L_{t}^{- 1} ∥ \geq ∥ e_{k_{0}} ∥_{X} / (∣ k_{0}^{2} - t c ∣ ∥ e_{k_{0}} ∥_{Y}) \to \infty$ as $t \to t_{0}$ .

3.2step 1.2step 1.3step 2.1F1givenalgebra

Estimate in the coercive range. For $t c < 1$ , $m \geq 1 - t c$ in step 1.3, so the sup and Hölder bounds there control $∥ u ∥_{C^{0, α}}$ and $∥ u^{'} ∥_{\infty}$ by $C (1 - t c)^{- 1} ∥ f ∥_{C^{0, α}}$ . From step 2.1, $u^{''} = - f - t c u$ , hence $∥ u^{''} ∥_{C^{0, α}} \leq C (1 - t c)^{- 1} ∥ f ∥_{C^{0, α}}$ . Thus $∥ u ∥_{C^{2, α}} \leq C (1 - t c)^{- 1} ∥ f ∥_{C^{0, α}}$ . Uniqueness follows from step 1.2; in particular $L_{t}$ is bijective for every non-spectral parameter, while this is the stated quantitative estimate on the coercive range.

4.1step 1.2step 2.1step 3.1F5givenalgebra

The two continuity properties of the bijectivity set. By steps 1.2 and 2.1, $I = [0, 1] ∖ {k^{2} / c : k^{2} \leq c}$ is exactly the bijectivity set. Its complement is finite, so $I$ is open in $[0, 1]$ . Every subinterval $J \subseteq [0, 1]$ on which all $L_{t}$ are injective contains no spectral parameter by step 1.2, hence $I \cap J = J$ is relatively closed in $J$ . These are the two properties inspected in the abstract method of continuity [F5]. At a spectral parameter the inverse norms on neighboring bijective parameters blow up as in step 3.1, so no a priori estimate uniform across that parameter can hold.

5.1step 1.2step 2.1step 3.1step 3.2step 4.1F5F6∎

Conclusion. The model family $L_{t} u = - u^{''} - t c u$ on $(0, π)$ is bounded $X \to Y$ for every $t$ , has the bijective base point $L_{0} = - d^{2} / d x^{2}$ , and has bijectivity set $I = {t : t c \notin {k^{2} : k \geq 1}}$ . For $t c < 1$ the eigenfunction expansion gives the inverse bound $C (1 - t c)^{- 1}$ ; at $t c = k_{0}^{2}$ the kernel is $span {\sin (k_{0} x)}$ and the range is the closed hyperplane orthogonal to it. Openness and the relative-closedness property hold by direct inspection of the finite exceptional set. This one-dimensional example illustrates the abstract method of continuity [F5] and its PDE-level application [F6].

Remarks

The example isolates the two ingredients of the method of continuity: a uniform inverse bound holds on compact parameter sets a positive distance from the spectrum; an open interval can avoid resonance while approaching it, in which case the inverse norm still diverges, and the base point $t = 0$ is bijective. The exceptional parameters are the zeros of $k^{2} - t c$ , where the inverse norm blows up like $1 / ∣ k^{2} - t c ∣$ .
The coefficient decay $∣ f_{k} ∣ \leq C ∥ f ∥_{C^{0, α}} k^{- α}$ is the only analytic input; it is exactly what makes $\sum k^{- 1 - α}$ and the splitting estimate for the H"older seminorm of $u$ converge, and this absolute-summability argument does not apply at $α = 0$ . For continuous forcing off resonance, direct integration of the ODE is an alternative route.

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

Bounded measurable coefficients do not give Schauder estimates

Statement refuted

The assertion that uniform ellipticity with merely bounded (even bounded continuous) principal coefficients suffices for a $C^{2, α}$ conclusion for classical $C^{2}$ solutions of $L u = f$ is false; the counterexample below exhibits a bounded continuous uniformly elliptic $L$ , a classical $C^{2}$ solution of $L u = 1$ , and a coefficient that fails to make $D^{2} u$ $α$ -Hölder for any exponent larger than the coefficient modulus. This does not contradict the interior Schauder theorem of this page, which assumes $C^{0, α}$ principal coefficients.

Facts & Assumptions

Given: $n \geq 2$ , $0 < β < α < 1$ , the unit ball $B_{1} (0) \subseteq R^{n}$ , the diagonal matrix field $A (x) = diag (1 + ∣ x_{1} ∣^{β}, 1, \dots, 1)$ , and the function $u (x) = \int_{0}^{x_{1}} \frac{x_{1} - s}{1 + ∣ s ∣^{β}} d s$ .

[A1]

The only choice assumption is Countable Choice ${A C}_{ω}$ ; no full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Uniform ellipticity and the nondivergence operator were fixed in Uniformly elliptic nondivergence-form operators and their frozen coefficients: $L = A^{i j} \partial_{i} \partial_{j}$ is uniformly elliptic with constants $λ, Λ$ when the symmetric matrix field satisfies $λ ∣ ξ ∣^{2} \leq A^{i j} (x) ξ_{i} ξ_{j} \leq Λ ∣ ξ ∣^{2}$ . The Hölder classes and their seminorms are those of Hölder spaces $C^{k, α}$ , closure and interior scaled norms, and $C^{k, α}$ domains. Coordinates $x_{i}$ and partials $\partial_{i}$ use the operator's one-based relabelling of coordinate $i - 1$ and $\partial_{i - 1}$ in $C^{k}$ maps and multi-index derivative notation in Euclidean space; multi-index derivatives retain that dependency's canonical order.

Counterexample

technique · direct

1.1F1givenalgebra

The coefficient field. For $x \in B_{1} (0)$ the matrix $A (x) = diag (1 + ∣ x_{1} ∣^{β}, 1, \dots, 1)$ is symmetric with eigenvalues $1 + ∣ x_{1} ∣^{β} \in [1, 2]$ , so it is bounded, continuous and uniformly elliptic on $B_{1} (0)$ with $λ = 1$ , $Λ = 2$ ; its off-diagonal entries vanish, and its $C^{0, α}$ seminorm on $B_{1} (0)$ is infinite because $∣ x_{1} ∣^{β}$ is not $α$ -Hölder at $0$ when $α > β$ .

1.2F1givenalgebra

The solution. Since the integrand $\frac{x_{1} - s}{1 + ∣ s ∣^{β}}$ vanishes at $s = x_{1}$ , differentiating under the integral sign gives $u^{'} (x_{1}) = \int_{0}^{x_{1}} \frac{d s}{1 + ∣ s ∣^{β}}$ and $u^{''} (x_{1}) = \frac{1}{1 + ∣ x_{1} ∣^{β}}$ ; both are continuous on $R$ , so $u \in C^{2}$ , and $u$ depends on $x_{1}$ only. Hence $A^{i j} (x) \partial_{i} \partial_{j} u (x) = A^{11} (x) u^{''} (x_{1}) = (1 + ∣ x_{1} ∣^{β}) \cdot \frac{1}{1 + ∣ x_{1} ∣^{β}} = 1$ for every $x \in B_{1} (0)$ , that is $L u = 1$ with $f \equiv 1$ , which is as smooth as possible.

2.1step 1.2F1algebra

Failure of the Hölder estimate. For $h \to 0$ one has $u^{''} (h) - u^{''} (0) = \frac{1}{1 + ∣ h ∣^{β}} - 1 = - \frac{∣ h ∣^{β}}{1 + ∣ h ∣^{β}}$ , so $\frac{∣ u^{''} (h) - u^{''} (0) ∣}{∣ h ∣^{α}} = \frac{∣ h ∣^{β - α}}{1 + ∣ h ∣^{β}} \to + \infty$ because $β < α$ ; hence $[u^{''}]_{0, α; B_{1} (0)} = + \infty$ and $u \notin C^{2, α} (B_{1} (0))$ for the given $α \in (0, 1)$ .

3.1step 1.1step 1.2step 2.1A1given∎

Conclusion. Steps 1.1, 1.2 and 2.1 give a uniformly elliptic nondivergence operator with bounded continuous (in particular bounded measurable) principal coefficients and a classical $C^{2}$ solution of $L u = 1$ on $B_{1} (0)$ whose second derivative fails to be $α$ -Hölder. Therefore bounded measurability of the coefficients does not force $C^{2, α}$ regularity; such a statement is false as stated, and the Hölder hypothesis on $A$ in the interior Schauder estimate of this page is not a technical convenience. The example uses no divergence-form interpretation and no choice beyond [A1].

Remarks

The failure is driven by the coefficient's own modulus: $A^{11} - 1 = ∣ x_{1} ∣^{β}$ is $β$ -Hölder, and the solution inherits exactly that modulus in $u^{''}$ ; a $C^{0, α}$ coefficient with $α > β$ would require $u^{''}$ to gain that Hölder regularity, which the example shows cannot be expected without the hypothesis.
The coefficient field here is continuous, so the example also refutes the stronger claim with "continuous" in place of "measurable"; its coefficient modulus is Dini and has vanishing mean oscillation as well. Those weaker hypotheses cannot force the particular $C^{2, α}$ conclusion refuted here; they may support different regularity conclusions.

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

Freezing cannot absorb a fixed oscillation on arbitrarily small balls

Statement refuted

Freezing coefficients is a stable device only when the coefficient oscillation at small scales actually vanishes. The assertion that the Hölder (or continuity) hypothesis on the principal coefficients in the freezing and Schauder arguments can be replaced by mere boundedness, or that the freezing radius alone can make an arbitrary fixed oscillation absorbable, is false: for a jump coefficient the freezing error is not even $α$ -Hölder at any scale, so no choice of the freezing radius makes the error term absorbable by the scaled norm.

Facts & Assumptions

Given: $n \geq 1$ , real numbers $a_{0} > b > 0$ , the coefficient field $a (x) : = a_{0} + b sign (x_{1})$ on $R^{n}$ , the diagonal matrix field $A (x) : = a (x) I_{n}$ (or, in $n = 1$ , the scalar coefficient), and centres $x_{0}$ with $x_{0, 1} = 0$ .

[F1]

The freezing estimate Freezing coefficients makes the Schauder error absorbable on a small ball requires $[A]_{0, α; B_{R} (x_{0})} \leq K$ and produces a bound $ρ^{2} ∥ (L - L_{0}) u ∥_{0, α; B_{ρ} (x_{0})}^{*} \leq ε ∥ u ∥_{2, α; B_{ρ} (x_{0})}^{*} + C_{ε} \sup ∣ u ∣$ , where $∥ g ∥_{0, α}^{*} = \sup ∣ g ∣ + ρ^{α} [g]_{0, α}$ ; in particular the left-hand side must be finite. The local Hölder seminorms are those of Local Hölder and scaled C-two-alpha norms on balls, the ballistic Euclidean balls those of Euclidean spheres and closed balls as subspaces of $R^{n}$ , and the nondivergence operator convention that of Uniformly elliptic nondivergence-form operators and their frozen coefficients.

Counterexample

technique · direct

1.1F1givenalgebra

The coefficient field. The function $sign (x_{1})$ is bounded and measurable (it is piecewise constant with a single jump), so $a$ is bounded and measurable, and $A = a I_{n}$ is a bounded symmetric measurable matrix field. For every $x$ , $a (x) \in [a_{0} - b, a_{0} + b]$ (with the standard convention $sign (0) = 0$ if used), and off the interface it equals one of the endpoint values. Thus the eigenvalues of $A (x)$ lie in $[a_{0} - b, a_{0} + b]$ , so $A$ is uniformly elliptic with $λ = a_{0} - b > 0$ and $Λ = a_{0} + b$ ; no continuity or Hölder regularity is available at $x_{1} = 0$ .

2.1step 1.1F1givenalgebra

The oscillation is fixed and never small. Let $x_{0}$ satisfy $x_{0, 1} = 0$ and let $ρ > 0$ . The two points $x_{0} \pm \frac{ρ}{2} e_{1}$ lie in $B_{ρ} (x_{0})$ and have first coordinate relative to the interface equal to $\pm ρ / 2$ , so $a$ takes both values $a_{0} + b$ and $a_{0} - b$ on the ball: ${osc}_{B_{ρ} (x_{0})} a = 2 b$ , independent of $ρ$ . Consequently the freezing modulus of continuity $ϵ (δ) : = \sup_{∣ x - y ∣ \leq δ} ∣ a (x) - a (y) ∣$ satisfies $ϵ (δ) = 2 b$ for every $δ > 0$ , because pairs straddling the hyperplane $x_{1} = 0$ are at arbitrarily small distance.

3.1step 2.1F1algebra

The freezing error is not Hölder at any scale. Let $0 < α < 1$ and consider the pair $x_{0} + t e_{1}$ , $x_{0} - t e_{1}$ with $t > 0$ : both lie in $B_{2 t} (x_{0})$ and $∣ a (x_{0} + t e_{1}) - a (x_{0} - t e_{1}) ∣ = 2 b$ , so $\frac{∣ a (x_{0} + t e_{1}) - a (x_{0} - t e_{1}) ∣}{∣ (x_{0} + t e_{1}) - (x_{0} - t e_{1}) ∣^{α}} = \frac{2 b}{(2 t)^{α}} ⟶ + \infty (t ↓ 0) .$ Hence $[a]_{0, α; B_{ρ} (x_{0})} = + \infty$ for every $ρ > 0$ and every $α \in (0, 1)$ : the coefficient is bounded but nowhere near Hölder on any ball centred on its interface.

4.1step 2.1step 3.1F1givenalgebra

No radius absorbs the frozen error. Take $L : = a (x) Δ$ with frozen part $L_{0} : = a_{0} Δ$ and $u (x) : = x_{1}^{2} / 2$ , which is $C^{\infty}$ with $Δ u = 1$ and finite scaled norm $∥ u ∥_{2, α; B_{ρ} (x_{0})}^{*} \leq C (1 + ρ + ρ^{2})$ on every ball of radius $ρ \leq 1$ . Then $(L - L_{0}) u = (a - a_{0}) Δ u = a - a_{0}$ on $B_{ρ} (x_{0})$ (up to the measure-zero hyperplane), so $ρ^{2} ∥ (L - L_{0}) u ∥_{0, α; B_{ρ} (x_{0})}^{*} \geq ρ^{2 + α} [a - a_{0}]_{0, α; B_{ρ} (x_{0})} = + \infty$ by step 3.1, while the right-hand side $ε ∥ u ∥_{2, α; B_{ρ} (x_{0})}^{*} + C_{ε} \sup_{B_{ρ}} ∣ u ∣$ is finite for every $ε > 0$ and every finite $C_{ε}$ . Hence the freezing estimate cannot hold for this coefficient for any choice of the freezing radius $ρ$ , and no radius can make the coefficient-oscillation term absorbable.

5.1step 1.1step 4.1given∎

Conclusion. A bounded measurable (indeed piecewise constant) uniformly elliptic coefficient with a fixed jump has oscillation $2 b$ at every scale and freezing error of infinite $α$ -Hölder seminorm; the freezing and Schauder arguments therefore genuinely need the vanishing small-scale oscillation provided by $C^{0, α}$ (or continuity) hypotheses, and this is not a technical convenience. The statement above is refuted, while the freezing lemma with its stated $C^{0, α}$ hypothesis is untouched by this example.

Remarks

The example also shows that in dimension one the coefficient $sign$ is the sharp obstruction: the jump in $a$ has size $2 b$ , and its one-dimensional distributional derivative is $2 b δ_{0}$ . The sign function itself is not a derivative of the Heaviside function.

Schauder and LpLp Elliptic Estimates — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The Schauder estimate on a quadratic Poisson solution: radius powers balance

Example

Facts & Assumptions

Verification

Remarks

A non-Dini continuous Poisson source can destroy the continuity of the second derivatives

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The Schauder estimate fails at the H"older endpoint α=1α=1

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The Riesz-transform formula for second derivatives of the Laplacian

Example

Facts & Assumptions

Verification

Remarks

Boundary W2,pW2,p regularity needs more than Lipschitz boundary

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The method of continuity on a constant-coefficient one-dimensional path

Example

Facts & Assumptions

Verification

Remarks

Bounded measurable coefficients do not give Schauder estimates

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Freezing cannot absorb a fixed oscillation on arbitrarily small balls

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

Schauder and $L^{p}$ Elliptic Estimates — Examples

The Schauder estimate fails at the H"older endpoint $α = 1$

Boundary $W^{2, p}$ regularity needs more than Lipschitz boundary