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Lax--Milgram and Weak Elliptic Solutions — Examples

1 · Prerequisites

2 · Summary

These companions compute, sharpen and stress-test the solvability theory of the main page. The sharp Dirichlet Poincar'e inequality on an interval and its sine witness give the exact constants used throughout, and the Neumann kernel is shown to be the span of the componentwise constants, motivating the mean-zero formulation and the per-component compatibility condition. On the interval the weak Poisson problem is solved explicitly by the Green function $G (x, y) = \min (x, y) - x y$ , and a one-dimensional form attains the $1 / α$ bound for the Lax--Milgram solution operator, so that estimate cannot be improved. Two counterexamples separate boundedness from coercivity: the zero form is solvable for no nonzero datum while the Neumann energy form fails coercivity on all of $H^{1}$ , and a large adverse zero-order term destroys Dirichlet coercivity at the sharp threshold $c = π^{2}$ , with explicit nonuniqueness at the endpoint. Coercivity is shown to be independent of symmetry through a nonsymmetric complex form and a drift operator solved by Lax--Milgram, and complex sesquilinear coercivity is contrasted with bilinear positivity, isolating the conjugation convention. Finally, arbitrary $L^{2}$ boundary data need not lie in the trace range $H^{1 / 2}$ and hence need not admit an $H^{1}$ lifting.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

The weak Dirichlet Poisson problem on an interval

Example

Assume the Axiom of Choice inherited through the cited suppliers, together with Countable Choice. Let $I = (0, 1)$ and $f \in L^{2} (I)$ . Define $u (x) : = \int_{0}^{1} G (x, y) f (y) d y, G (x, y) : = \min (x, y) - x y .$ Then $G$ is continuous on $[0, 1]^{2}$ with $G (0, \cdot) = G (1, \cdot) = 0$ , $u \in H_{0}^{1} (I)$ is the unique weak solution of $- u^{''} = f$ with zero boundary values in the sense of Existence and uniqueness for the weak Dirichlet Poisson problem, and $\int_{0}^{1} u^{'} (x) \overline{φ^{'} (x)} d x = \int_{0}^{1} f (x) \overline{φ (x)} d x for every φ \in C_{c}^{\infty} (I) .$ The solution is recovered by two integrations: $u^{'}$ is absolutely continuous, $u^{''} = - f$ a.e. and $u (0) = u (1) = 0$ ; for $f = 1$ this gives the explicit $u (x) = \frac{1}{2} x (1 - x)$ . More generally, if $F \in H^{- 1} (I)$ is represented as $F (v) = (f_{0}, v)_{L^{2}} + (f_{1}, v^{'})_{L^{2}}$ with $f_{0}, f_{1} \in L^{2} (I)$ (Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence), then the weak solution is $u (x) = \int_{0}^{1} G (x, y) f_{0} (y) d y + \int_{0}^{1} \partial_{y} G (x, y) f_{1} (y) d y$ , where $\partial_{y} G$ is taken in the distributional sense, matching the one-dimensional integration-by-parts formula. This illustrates item 13 of the design on a one-dimensional model; no general Green-function theory is claimed.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the interval $I = (0, 1)$ ; a class $f \in L^{2} (I)$ ; the Green kernel $G (x, y) = \min (x, y) - x y$ on $[0, 1]^{2}$ ; and $u (x) : = \int_{0}^{1} G (x, y) f (y) d y$ .

[F1]

Kernel facts: $G$ is continuous on $[0, 1]^{2}$ with $G (0, \cdot) = G (1, \cdot) = 0$ ; for $y \neq x$ one has $\partial_{x} G (x, y) = 1_{y > x} - y \in [- 1, 1]$ and $\partial_{y} G (x, y) = 1_{x > y} - x \in [- 1, 1]$ , so both partial derivatives are bounded by $1$ in modulus (direct computation from $G = \min (x, y) - x y$ ; The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ supplies the underlying linearity and Integral over a measurable subset the Lebesgue integrals).

[F2]

Dominated convergence gives convergence of integrals under an integrable bound (Dominated convergence). An indefinite integral of an $L^{1}$ function is absolutely continuous with derivative equal to the integrand a.e. by The indefinite integral of an $L^{1}$ function is absolutely continuous and The indefinite integral of an $L^{1}$ function is differentiable almost everywhere, while the recovery formula for an absolutely continuous function is Fundamental theorem of calculus for absolutely continuous functions. Integration by parts for absolutely continuous factors is Integration by parts for absolutely continuous functions, applied componentwise over $C$ . Products of absolutely continuous functions remain absolutely continuous (The product of two absolutely continuous functions is absolutely continuous); Absolute continuity of the integral controls integrals on the shrinking boundary strips, and Holder's inequality for integrals, including the endpoint cases bounds the products.

[F3]

Membership criterion proved below: an absolutely continuous $w$ on $[0, 1]$ with $w (0) = w (1) = 0$ and $w^{'} \in L^{2} (0, 1)$ lies in $H_{0}^{1} (I)$ . The boundary cutoff uses the standard smooth step $σ$ , which takes values in $[0, 1]$ and has bounded derivative; its product and chain rules give a compactly supported $W^{1, 2}$ approximation. The approximation is then zero-extended and mollified, using the ACL characterisation, the compact-support zero-extension theorem, convergence of mollifiers, and the density definition of $H_{0}^{1}$ (The standard smooth step function, Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The ACL characterisation of $W^{1, p}$ , Compactly supported Sobolev functions extend by zero in every integer order, The mollifier family generated by a unit-mass smooth bump, A unit-mass smooth bump generates an $L^{1}$ approximate identity, Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ , Convolution with a mollifier is smooth, and derivatives pass under the integral sign, Zero-boundary Sobolev space as a norm closure, One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives).

[F4]

Lax--Milgram existence and uniqueness on $I$ : for every $F \in H^{- 1} (I)$ there is a unique $u \in H_{0}^{1} (I)$ with $\int_{0}^{1} u^{'} \overline{v^{'}} = F (v)$ for all $v \in H_{0}^{1} (I)$ (Existence and uniqueness for the weak Dirichlet Poisson problem, The negative Sobolev space $H^{- 1} (Ω)$ ); if $F (v) = (f, v)_{L^{2}}$ with $f \in L^{2}$ , this datum lies in $H^{- 1}$ (Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence, The space $L^{p} (μ)$ as the quotient by null functions, Real and imaginary parts, complex conjugation, and modulus).

Proof

1.1F1F2

Differentiation under the integral: fixing $x$ and letting $h \to 0$ , the kernel identity $(G (x + h, y) - G (x, y)) / h \to \partial_{x} G (x, y)$ holds for every $y \neq x$ , hence for almost every $y$ , and the quotients are bounded by $1$ because $G$ is $1$ -Lipschitz in its first variable on $[0, 1]$ ; since $∣ f ∣ \in L^{1}$ , dominated convergence gives $u^{'} (x) = \int_{0}^{1} \partial_{x} G (x, y) f (y) d y = \int_{x}^{1} f (y) d y - \int_{0}^{1} y f (y) d y .$

1.2F3algebra

Boundary cutoff: prove the criterion of [F3]. Let $w$ be absolutely continuous on $[0, 1]$ with $w (0) = w (1) = 0$ and $w^{'} \in L^{2}$ . For $m \geq 4$ set $χ_{m} (x) : = σ (m x - 1) σ (m (1 - x) - 1)$ and $w_{m} : = χ_{m} w$ . Then $χ_{m}$ vanishes on $[0, 1 / m] \cup [1 - 1 / m, 1]$ , equals $1$ on $[2 / m, 1 - 2 / m]$ , takes values in $[0, 1]$ , and $∣ χ_{m}^{'} ∣ \leq 2 m C_{σ}$ by the chain and product rules. The product $w_{m}$ is absolutely continuous with derivative $χ_{m}^{'} w + χ_{m} w^{'} \in L^{2}$ and compact support in $I$ , so $w_{m} \in W^{1, 2} (I)$ .

2.1F1F2step 1.1

Regularity and boundary values: the formula of step 1.1 exhibits $u^{'}$ as the sum of the continuous function $x \mapsto \int_{x}^{1} f$ and a constant, so $u \in C^{1} ([0, 1])$ ; $u (0) = \int G (0, y) f = 0$ and $u (1) = 0$ by [F1], and the fundamental theorem gives $u (x) = \int_{0}^{x} u^{'} (t) d t$ . Moreover $u^{'}$ is the difference of an absolutely continuous function and a constant, so $u^{''} = - f$ almost everywhere.

2.2F3step 1.2algebra

Convergence in $H^{1}$ : put $E_{m} : = (0, 2 / m) \cup (1 - 2 / m, 1)$ . Since $w_{m} = w$ off $E_{m}$ and $w \in L^{2}$ , $∥ w_{m} - w ∥_{L^{2}} \to 0$ ; in $L^{2}$ , $(w_{m} - w)^{'} = (χ_{m} - 1) w^{'} + χ_{m}^{'} w$ , and the first term tends to zero because $w^{'} \in L^{2}$ and the strips shrink to the endpoints. Near $0$ , $∣ w (x) ∣^{2} = ∣ \int_{0}^{x} w^{'} (t) d t ∣^{2} \leq x \int_{0}^{x} ∣ w^{'} (t) ∣^{2} d t$ , while near $1$ , $∣ w (x) ∣^{2} = ∣ \int_{x}^{1} w^{'} (t) d t ∣^{2} \leq (1 - x) \int_{x}^{1} ∣ w^{'} (t) ∣^{2} d t$ . Therefore $\int_{1 / m}^{2 / m} ∣ χ_{m}^{'} w ∣^{2} d x \leq 6 C_{σ}^{2} \int_{0}^{2 / m} ∣ w^{'} ∣^{2} d t and \int_{1 - 2 / m}^{1 - 1 / m} ∣ χ_{m}^{'} w ∣^{2} d x \leq 6 C_{σ}^{2} \int_{1 - 2 / m}^{1} ∣ w^{'} ∣^{2} d t,$ and both bounds tend to zero. Thus $∥ w_{m} - w ∥_{H^{1} (0, 1)} \to 0$ .

3.1F3step 2.2

Mollification and closure: each $w_{m}$ has compact support inside $I$ , so its zero extension $W_{m}$ belongs to $W^{1, 2} (R)$ by [F3]. Mollifying at sufficiently small scales gives $W_{m} * ρ_{ε} \in C_{c}^{\infty} (I)$ . The weak-derivative identity with test $y \mapsto ρ_{ε} (x - y)$ gives $(W_{m} * ρ_{ε})^{'} = W_{m}^{'} * ρ_{ε}$ ; applying the $L^{2}$ approximate-identity result separately to $W_{m}$ and $W_{m}^{'}$ proves convergence in $W^{1, 2}$ . Hence $w_{m} \in H_{0}^{1} (I)$ . Since this space is closed and $w_{m} \to w$ in $H^{1}$ , $w \in H_{0}^{1} (I)$ . Applying the criterion to step 2.1 gives $u \in H_{0}^{1} (I)$ .

3.2F2step 2.1

Weak identity on test functions: for $φ \in C_{c}^{\infty} (I)$ , integration by parts on $[0, 1]$ and step 2.1 give $\int_{0}^{1} u^{'} \overline{φ^{'}} d x = [u^{'} \overline{φ}]_{0}^{1} - \int_{0}^{1} u^{''} \overline{φ} d x = \int_{0}^{1} f \overline{φ} d x,$ the boundary term vanishing because $φ$ has compact support in $I$ and $u^{''} = - f$ a.e.

4.1F4step 3.1step 3.2algebra

Identification and uniqueness: the right-hand side $v \mapsto (f, v)_{L^{2}}$ is an element of $H^{- 1} (I)$ by [F4], and both sides of the identity are bounded in $v$ on $H_{0}^{1} (I)$ by H"older; since $C_{c}^{\infty} (I)$ is dense in $H_{0}^{1} (I)$ , the identity of step 3.2 extends to every $v \in H_{0}^{1} (I)$ . Hence $u$ is the unique weak solution of $- u^{''} = f$ with zero boundary values. For $f = 1$ the formula gives $u^{'} (x) = (1 - x) - \frac{1}{2} = \frac{1}{2} - x$ and $u (x) = \frac{1}{2} x (1 - x)$ .

5.1F4step 3.1step 4.1algebra

General $H^{- 1}$ datum: let $F (v) = (f_{0}, v)_{L^{2}} + (f_{1}, v^{'})_{L^{2}}$ with $f_{0}, f_{1} \in L^{2} (I)$ and define $w (x) : = \int_{0}^{1} \partial_{y} G (x, y) f_{1} (y) d y = \int_{0}^{x} f_{1} (y) d y - x \int_{0}^{1} f_{1} (y) d y$ . Then $w$ is absolutely continuous with $w (0) = w (1) = 0$ and $w^{'} = f_{1} - \int_{0}^{1} f_{1} \in L^{2}$ , so $w \in H_{0}^{1} (I)$ by the criterion of step 3.1; for $v \in H_{0}^{1} (I)$ one computes $\int_{0}^{1} w^{'} \overline{v^{'}} = \int_{0}^{1} f_{1} \overline{v^{'}} - (\int_{0}^{1} f_{1}) \overline{\int_{0}^{1} v^{'}}$ , and $\int_{0}^{1} v^{'} = 0$ for $v \in H_{0}^{1}$ , by approximating $v$ in $H^{1}$ by compactly supported smooth $v_{j}$ , for which $\int v_{j}^{'} = 0$ , and using $∣ \int (v^{'} - v_{j}^{'}) ∣ \leq ∥ v^{'} - v_{j}^{'} ∥_{2}$ , so the identity $\int_{0}^{1} w^{'} \overline{v^{'}} = \int_{0}^{1} f_{1} \overline{v^{'}}$ holds. Combined with step 4.1 applied to $f_{0}$ , the function $u = u_{0} + w$ with $u_{0} (x) = \int G (x, y) f_{0} (y) d y$ satisfies the weak equation for the datum $F$ , and by uniqueness it is the weak solution; this is the displayed Green representation with $\partial_{y} G$ acting on $f_{1}$ .

6.1step 4.1step 5.1∎

Conclusion: the Green function representation produces the unique weak solution on the interval, with the weak identity and the explicit case $f = 1$ giving $u (x) = \frac{1}{2} x (1 - x)$ , and the general divergence-form datum is handled by the same kernel with $\partial_{y} G$ ; no general Green-function theory is claimed.

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$L^{2}$ forcing defines an $H^{- 1}$ functional

Example

Assume the Axiom of Choice inherited through the cited suppliers, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open, nonempty and bounded in one direction, with Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction). For $f \in L^{2} (Ω)$ define $F_{f} (v) : = (f, v)_{L^{2}}$ , $v \in H_{0}^{1} (Ω)$ . Then $F_{f} \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ) with $∥ F_{f} ∥_{H^{- 1}} \leq C_{P} ∥ f ∥_{L^{2}} .$ If moreover $f \in H_{0}^{1} (Ω)$ and $f \neq 0$ , testing $v = f$ gives the data-dependent lower bound $∥ F_{f} ∥_{H^{- 1}} \geq \frac{∥ f ∥_{L^{2}}^{2}}{∥ f ∥_{H_{0}^{1}}};$ no uniform positive lower bound by $∥ f ∥_{L^{2}}$ holds on all of $H_{0}^{1}$ , as the interval sine sequence below shows. The map $f \mapsto F_{f}$ is injective: if $F_{f} = 0$ then $(f, v)_{L^{2}} = 0$ for all $v \in C_{c}^{\infty} (Ω)$ , hence $f = 0$ a.e. Taking $f = 1$ on a bounded interval $(0, L)$ shows that the upper bound $C_{P} ∥ f ∥_{L^{2}}$ is at least $(L / π) \sqrt{L}$ there and grows with the domain diameter. This is the exact cheap estimate that the plan records as consumed by the weak Dirichlet problem ( $L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound); it is not surjectivity of the embedding, which fails for $n \geq 1$ (an explicit datum outside its range is constructed in step 1.4; Every $H^{- 1}$ functional is an $L^{2}$ function plus a divergence supplies the general representation).

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open, nonempty $Ω \subseteq R^{n}$ bounded in one direction with Poincar'e constant $C_{P}$ for $W_{0}^{1, 2}$ ; a class $f \in L^{2} (Ω)$ ; the functional $F_{f} (v) = (f, v)_{L^{2}}$ on $H_{0}^{1} (Ω)$ .

[F1]

Upper estimate: the functional $v \mapsto (f, v)_{L^{2}}$ is conjugate-linear, well defined on classes, and $∥ F_{f} ∥_{H^{- 1}} \leq C_{P} ∥ f ∥_{L^{2}}$ ; this is $L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound with $f_{0} = f$ and $f_{1} = \dots = f_{n} = 0$ , using $∥ v ∥_{L^{2}} \leq C_{P} ∥ D v ∥_{L^{2}}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F2]

$H^{- 1}$ norm and the lower bound: $∥ F ∥_{H^{- 1}} = \sup_{∥ v ∥_{H_{0}^{1}} \leq 1} ∣ F (v) ∣$ , so for $f \neq 0$ testing with $v = f \in H_{0}^{1}$ gives $∥ F_{f} ∥_{H^{- 1}} \geq ∣ (f, f)_{L^{2}} ∣ / ∥ f ∥_{H_{0}^{1}} = ∥ f ∥_{L^{2}}^{2} / ∥ f ∥_{H_{0}^{1}}$ . Poincar'e controls $∥ f ∥_{L^{2}}$ by $∥ D f ∥_{L^{2}}$ and supplies no upper bound of $∥ f ∥_{H_{0}^{1}}$ by $∥ f ∥_{L^{2}}$ (The negative Sobolev space $H^{- 1} (Ω)$ , Integer-order Sobolev spaces and their norms, The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction, Zero-boundary Sobolev space as a norm closure).

[F3]

Injectivity input: the map $f \mapsto u_{f}$ , $u_{f} (φ) = \int_{Ω} f φ$ , is an injection from $L_{l o c}^{1} (Ω)$ modulo almost-everywhere equality into the distributions $D^{'} (Ω)$ , $D^{'} (Ω)$ being the continuous linear functionals on the test-function space $D (Ω) = C_{c}^{\infty} (Ω)$ ; hence $u_{f} = 0$ forces $f = 0$ a.e. (Locally integrable functions embed in distributions, Regular distribution from a locally integrable function, Test function space d of an open set). Every $L^{2}$ class on $Ω$ lies in $L_{l o c}^{1} (Ω)$ , because $\int_{K} ∣ f ∣ \leq ∣ K ∣^{1 / 2} ∥ f ∥_{L^{2} (K)}$ on each compact $K \subseteq Ω$ (Holder's inequality for integrals, including the endpoint cases).

[F4]

Sharp interval inequality: on $Ω = (0, L)$ the function $ϕ (x) = \sin (π x / L)$ lies in $H_{0}^{1} (0, L)$ , is nonzero, and satisfies $∥ ϕ ∥_{L^{2}} = (L / π) ∥ ϕ^{'} ∥_{L^{2}}$ , so every constant $C$ admissible in $∥ v ∥_{L^{2}} \leq C ∥ \nabla v ∥_{L^{2}}$ on $H_{0}^{1} (0, L)$ satisfies $C \geq L / π$ (The sharp Dirichlet Poincare inequality on an interval).

[F5]

The constant function $1$ on $(0, L)$ has $∥ 1 ∥_{L^{2} (0, L)} = \sqrt{L}$ : the constant $1$ is bounded and Riemann integrable on $[0, L]$ with Riemann integral $L$ , and a bounded Riemann integrable function on a closed bounded interval has the same Lebesgue integral (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, The space $L^{p} (μ)$ as the quotient by null functions, Integral over a measurable subset).

[F6]

For integers $k \geq 1$ , $f_{k} (x) = \sin (k π x)$ on $(0, 1)$ satisfies $∥ f_{k} ∥_{L^{2}}^{2} = 1 / 2$ , $∥ f_{k}^{'} ∥_{L^{2}}^{2} = (k π)^{2} / 2$ , and $f_{k}^{''} = - (k π)^{2} f_{k}$ by the trigonometric identities and derivative rules (The derivatives of sine and cosine are cosine and minus sine, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F7]

The standard smooth step $σ$ takes values in $[0, 1]$ , equals $0$ on $(- \infty, 0]$ and $1$ on $[1, \infty)$ , and has bounded derivative $C_{σ} : = \sup ∣ σ^{'} ∣ < \infty$ because $σ^{'}$ is continuous and vanishes outside the compact interval $[0, 1]$ (The standard smooth step function, Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value).

[F8]

Integration by parts applies to smooth compactly supported test functions on $I$ , and $C_{c}^{\infty} (I)$ is dense in $H_{0}^{1} (I)$ by its definition as a Sobolev closure; the pairings in the identity are continuous in the $H_{0}^{1}$ norm (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ , Zero-boundary Sobolev space as a norm closure, Holder's inequality for integrals, including the endpoint cases).

[F9]

Arbitrary $L^{2}$ data $f_{0}, \dots, f_{n}$ define a bounded functional $F (v) = (f_{0}, v) + \sum_{i} (f_{i}, D_{i} v)$ by $L^{2}$ forcing and divergence data embed in $H^{- 1}$ with a quantitative bound. Smooth compactly supported bumps exist inside every ball, by A smooth bump between concentric Euclidean balls and translation; their products give bumps inside boxes. Fubini factors integrals of products on boxes (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability), and the fundamental theorem evaluates the smooth one-variable derivative integrals (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

Proof

1.1F1

Upper bound: [F1] applied with $f_{0} = f$ and all $f_{i} = 0$ gives $∥ F_{f} ∥_{H^{- 1}} \leq C_{P} ∥ f ∥_{L^{2}}$ , so $F_{f} \in H^{- 1} (Ω)$ .

1.2F2

Exact lower bound on the Sobolev space: if $f \in H_{0}^{1} (Ω)$ is nonzero, then $f$ is an admissible test function and $\frac{∣ F_{f} (f) ∣}{∥ f ∥_{H_{0}^{1}}} = \frac{∥ f ∥_{L^{2}}^{2}}{∥ f ∥_{H_{0}^{1}}}$ is a lower bound for $∥ F_{f} ∥_{H^{- 1}}$ , because that norm is the supremum of $∣ F_{f} (v) ∣ / ∥ v ∥_{H_{0}^{1}}$ over nonzero $v$ .

1.3F4F5algebra

Interval scaling: take $Ω = (0, L)$ and $f = 1$ . By the sharp interval inequality every admissible Poincar'e constant $C$ for $(0, L)$ satisfies $C \geq L / π$ , and $∥ f ∥_{L^{2} (0, L)} = \sqrt{L}$ ; hence the displayed bound $C_{P} ∥ f ∥_{L^{2}}$ is at least $(L / π) \sqrt{L}$ and grows with the diameter $L$ of the domain.

1.4F1F9algebra

The embedding is not onto. Choose a box $Q = (s - r, s + r) \times Q^{'}$ compactly contained in the nonempty open set $Ω$ , a real $ψ \in C_{c}^{\infty} (- 1, 1)$ with $ψ (0) = 1$ , and a nonzero real $η \in C_{c}^{\infty} (Q^{'})$ using [F9]. For $n = 1$ , omit the transverse factor and set $C = 1$ ; otherwise put $C : = \int_{Q^{'}} η^{2} > 0$ . Set $f_{1} (x) = 1_{(s, s + r)} (x_{1}) η (x^{'})$ on $Q$ and zero outside, with $f_{0} = f_{2} = \dots = f_{n} = 0$ . These are $L^{2}$ data, so $F (v) : = \int f_{1} \overline{D_{1} v}$ belongs to $H^{- 1}$ by [F9]. For $0 < ε < r$ , take $v_{ε} (x) = ψ ((x_{1} - s) / ε) η (x^{'}) \in C_{c}^{\infty} (Q)$ . Fubini and the fundamental theorem give $F (v_{ε}) = C (ψ (r / ε) - ψ (0)) = - C$ , whereas $∥ v_{ε} ∥_{2}^{2} = ε C \int_{- 1}^{1} ψ^{2} \to 0$ . If $F = F_{h}$ for some $h \in L^{2} (Ω)$ , H"older would imply $∣ F (v_{ε}) ∣ \leq ∥ h ∥_{2} ∥ v_{ε} ∥_{2} \to 0$ , contradicting $C > 0$ . Thus a datum outside the $L^{2}$ embedding exists on every such nonempty $Ω$ .

2.1F2F6F7F8step 1.1algebra

No uniform $L^{2}$ lower bound: on $I = (0, 1)$ , let $f_{k} (x) = \sin (k π x)$ . To see $f_{k} \in H_{0}^{1} (I)$ , for each fixed $k$ use $η_{m} (x) = σ (m x - 1) σ (m (1 - x) - 1)$ and set $g_{k, m} = η_{m} f_{k}$ . Then $g_{k, m} \in C_{c}^{\infty} (I)$ , $η_{m} = 1$ on $[2 / m, 1 - 2 / m]$ , $∣ η_{m}^{'} ∣ \leq 2 m C_{σ}$ , and on the two boundary strips $∣ f_{k} ∣ \leq 2 k π / m$ and $∣ f_{k}^{'} ∣ \leq k π$ . Thus $∥ g_{k, m} - f_{k} ∥_{L^{2}}^{2} \leq 16 (k π)^{2} / m^{3}$ and $∥ (g_{k, m} - f_{k})^{'} ∥_{L^{2}}^{2} \leq 4 (k π)^{2} (1 + 4 C_{σ})^{2} / m$ , so $g_{k, m} \to f_{k}$ in $H^{1}$ and $f_{k} \in H_{0}^{1} (I)$ . By [F8], first for $v \in C_{c}^{\infty} (I)$ and then by density for every $v \in H_{0}^{1} (I)$ , $F_{f_{k}} (v) = \int_{0}^{1} f_{k} \overline{v} d x = \frac{1}{(k π)^{2}} \int_{0}^{1} f_{k}^{'} \overline{v^{'}} d x .$ Cauchy--Schwarz and [F6] give $∥ F_{f_{k}} ∥_{H^{- 1}} \leq ∥ f_{k} ∥_{L^{2}} / (k π)$ , so $∥ F_{f_{k}} ∥_{H^{- 1}} / ∥ f_{k} ∥_{L^{2}} \to 0$ . Therefore no uniform positive lower bound by $∥ f ∥_{L^{2}}$ holds on all of $H_{0}^{1} (I)$ .

3.1F3step 1.1step 1.2step 2.1

Injectivity: if $F_{f} = 0$ , then $(f, v)_{L^{2}} = 0$ for every $v \in C_{c}^{\infty} (Ω) \subseteq H_{0}^{1} (Ω)$ . Given $φ \in C_{c}^{\infty} (Ω)$ , applying this to $v = \overline{φ}$ gives $u_{f} (φ) = \int_{Ω} f φ = 0$ , so $u_{f}$ is the zero distribution; the injectivity in [F3] then gives $f = 0$ a.e., that is, $f$ is the zero class. Hence $f \mapsto F_{f}$ is injective. When $f \in H_{0}^{1} (Ω)$ , steps 1.1 and 1.2 give an upper bound and the valid data-dependent lower bound; step 2.1 shows that this lower bound cannot be replaced by a uniform positive multiple of $∥ f ∥_{L^{2}}$ .

4.1F1step 2.1step 3.1step 1.4step 1.3step 1.2∎

Conclusion: $L^{2}$ forcing defines an element of $H^{- 1}$ with the quantitative upper bound of step 1.1, the valid data-dependent lower bound of step 1.2 for $f \in H_{0}^{1}$ , injectivity by step 3.1, and diameter growth of the upper bound by step 1.3. Step 2.1 shows why there is no uniform lower estimate in the $L^{2}$ norm; this example is not surjectivity of $L^{2} (Ω) ↪ H^{- 1} (Ω)$ , as the explicit datum of step 1.4 is outside its range.

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A nonsymmetric coercive elliptic form

Example

Assume the Axiom of Choice and Countable Choice. Let $n \geq 2$ , let $Ω \subseteq R^{n}$ be nonempty, open and bounded in one direction, and let $C_{P}$ be its Poincar'e constant for $W_{0}^{1, 2}$ . Set $b = e_{1}$ and define $a (u, v) : = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x + \int_{Ω} \partial_{1} u \overline{v} d x (u, v \in H_{0}^{1} (Ω)) .$ This is a bounded sesquilinear form with bound $2$ and is coercive with constant $1 / (1 + C_{P}^{2})$ . It is not symmetric: choose $x_{0} \in Ω$ with $x_{0, 2} \neq 0$ , a ball $B_{r} (x_{0}) \subset Ω$ , a nonzero real radial bump $η$ supported in that ball, and put $u = x_{2} η$ , $v = x_{1} η$ . Then $a (u, v) - \overline{a (v, u)} = - \int_{Ω} x_{2} η^{2} d x = - x_{0, 2} \int_{Ω} η^{2} d x \neq 0.$ Thus Lax--Milgram (The Lax--Milgram theorem) applies to this weak Dirichlet problem for $L u = - Δ u + \partial_{1} u$ , while the minimisation characterisation of Symmetric Lax--Milgram is energy minimisation does not apply. This is the companion example of Nonsymmetric Lax--Milgram is not a scalar minimisation principle and of the drift term in A large adverse zero-order term destroys Dirichlet coercivity.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; $n \geq 2$ ; a nonempty open $Ω \subseteq R^{n}$ bounded in one direction; $b = e_{1}$ ; the form $a$ above; and the stated bump supported in a ball about $x_{0}$ with $x_{0, 2} \neq 0$ .

[F1]

Since $b = e_{1}$ has $∥ b ∥_{L^{\infty}} = 1$ , Cauchy--Schwarz gives $∣ a (u, v) ∣ \leq ∥ \nabla u ∥_{2} ∥ \nabla v ∥_{2} + ∥ \nabla u ∥_{2} ∥ v ∥_{2} \leq 2 ∥ u ∥_{H_{0}^{1}} ∥ v ∥_{H_{0}^{1}}$ ; the form is bounded and sesquilinear (The elliptic form is well defined and bounded on $H^{1}$ , Uniformly elliptic divergence-form operators and their sesquilinear forms, Holder's inequality for integrals, including the endpoint cases).

[F2]

Poincar'e gives $∥ u ∥_{H_{0}^{1}}^{2} \leq (1 + C_{P}^{2}) ∥ \nabla u ∥_{2}^{2}$ , so the principal form has coercivity constant $1 / (1 + C_{P}^{2})$ (Coercivity of the principal Dirichlet form, The Sobolev space $H^{1}$ is a Hilbert space, Integer-order Sobolev spaces and their norms).

[F3]

For $φ \in C_{c}^{\infty} (Ω)$ , its zero extension is smooth and compactly supported in $R^{n}$ . Choose $R > 0$ so that its support lies in $(- R, R)^{n}$ . For each fixed $x^{'} = (x_{2}, \dots, x_{n})$ , the function $t \mapsto ∣ φ (t, x^{'}) ∣^{2}$ has compact support in $(- R, R)$ , so the one-dimensional fundamental theorem gives $\int_{- R}^{R} \partial_{1} ∣ φ (t, x^{'}) ∣^{2} d t = 0$ . Fubini on the cube then gives $\int_{Ω} \partial_{1} ∣ φ ∣^{2} = 0$ , hence $Re \int_{Ω} \partial_{1} φ \overline{φ} = 0$ . Also $C_{c}^{\infty} (Ω)$ is dense in $H_{0}^{1} (Ω)$ , and $u \mapsto \int_{Ω} \partial_{1} u \overline{u}$ is continuous in the $H^{1}$ norm by Cauchy--Schwarz and [F1] (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Test function space d of an open set, Zero-boundary Sobolev space as a norm closure, Holder's inequality for integrals, including the endpoint cases).

[F4]

The coordinate product rule gives $\partial_{1} (x_{2} η) = x_{2} \partial_{1} η$ and $\partial_{1} (x_{1} η) = η + x_{1} \partial_{1} η$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ). For a radial bump about $x_{0}$ , reflection $x_{2} \mapsto 2 x_{0, 2} - x_{2}$ leaves $η^{2}$ unchanged and has absolute Jacobian $1$ ; applying The published Riemann change-of-variables theorem already gives the Lebesgue formula for continuous compactly supported integrands on $R^{n}$ to $(x_{2} - x_{0, 2}) η^{2}$ shows its integral equals its negative, hence is zero.

[F5]

The standard smooth step $σ$ is smooth, takes values in $[0, 1]$ , vanishes for $t \leq 0$ and equals $1$ for $t \geq 1$ (The standard smooth step function). Repeated coordinate chain and product rules give smoothness of its composition with a polynomial (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F6]

The real or complex Hilbert space $H_{0}^{1} (Ω)$ is complete, and Lax--Milgram applies to every bounded coercive sesquilinear form without symmetry; the energy-minimisation conclusion requires symmetry (The Sobolev space $H^{1}$ is a Hilbert space, The Lax--Milgram theorem, Symmetric Lax--Milgram is energy minimisation, Nonsymmetric Lax--Milgram is not a scalar minimisation principle, Weak Dirichlet solutions for a divergence-form operator).

Proof

Given: The Axiom of Choice and Countable Choice; the stated $Ω$ , $b = e_{1}$ and form; and the bump $η$ .

1.1F1

Boundedness: by [F1] the form is bounded with $∣ a (u, v) ∣ \leq 2 ∥ u ∥_{H_{0}^{1}} ∥ v ∥_{H_{0}^{1}}$ and is linear in its first argument and conjugate-linear in its second.

1.2F3

The drift has zero real part: for $φ \in C_{c}^{\infty} (Ω)$ , [F3] gives $2 Re \int_{Ω} \partial_{1} φ \overline{φ} = \int_{Ω} \partial_{1} ∣ φ ∣^{2} = 0$ . By continuity and density in [F3], this extends to every $u \in H_{0}^{1} (Ω)$ , so $Re a (u, u) = ∥ \nabla u ∥_{2}^{2}$ .

1.3F3F4F5algebraconstruct

Nonsymmetry: openness and nonemptiness of $Ω$ give $x_{0} \in Ω$ with $x_{0, 2} \neq 0$ and a ball $B_{r} (x_{0}) \subset Ω$ . Set $s = 3 r / 4$ and $η (x) = σ ((s^{2} - ∣ x - x_{0} ∣^{2}) / (s^{2} - r^{2} / 4))$ . By [F5] this is a smooth real radial function, equals $1$ on ${\overline{B}}_{r / 2} (x_{0})$ and vanishes outside $B_{s} (x_{0})$ ; hence its support is contained in ${\overline{B}}_{s} (x_{0}) \subset B_{r} (x_{0})$ and $\int η^{2} > 0$ . Thus $u = x_{2} η$ and $v = x_{1} η$ are admissible smooth compactly supported tests. Their principal parts cancel, and [F4] gives $a (u, v) - \overline{a (v, u)} = \int_{Ω} (\partial_{1} u v - \partial_{1} v u) d x = - \int_{Ω} x_{2} η^{2} d x = - x_{0, 2} \int_{Ω} η^{2} d x \neq 0.$

2.1F1F2F6step 1.2step 1.3

Coercivity and solvability: by [F2] and step 1.2, $Re a (u, u) = ∥ \nabla u ∥_{2}^{2} \geq (1 + C_{P}^{2})^{- 1} ∥ u ∥_{H_{0}^{1}}^{2}$ , so the bounded form is coercive. Lax--Milgram gives the weak Dirichlet solution, while the minimisation result does not apply to this nonsymmetric form.

3.1F6step 2.1step 1.3∎

Conclusion: $b = e_{1}$ gives a concrete bounded, coercive, nonsymmetric form on every such nonempty open $Ω$ in dimension $n \geq 2$ ; the example demonstrates exactly why symmetry is required for the energy-minimisation characterization.

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A bounded form without coercivity need not be solvable

Statement refuted

Let $H$ be a nonzero real or complex Hilbert space and let $a \equiv 0$ , so $a$ is a bounded sesquilinear form with bound $M = 0$ , but $a$ is not coercive: $Re a (u, u) = 0$ for all $u$ , so no $α > 0$ satisfies $Re a (u, u) \geq α ∥ u ∥^{2}$ at any $u \neq 0$ . Let $F \neq 0$ be a bounded conjugate-linear functional on $H$ . Then the equation $a (u, v) = F (v)$ for all $v \in H$ has no solution, since its left side is identically $0$ while the right side is not. Hence boundedness alone does not imply existence or uniqueness, and the coercivity hypothesis of The Lax--Milgram theorem cannot be dropped. The same witness shows that the estimate $∥ u ∥ \leq ∥ F ∥ / α$ has no content without $α > 0$ .

Facts & Assumptions

Given: A nonzero real or complex Hilbert space $H$ ; the zero form $a \equiv 0$ ; and a nonzero bounded conjugate-linear functional $F$ on $H$ .

[F1]

$a \equiv 0$ is sesquilinear and bounded with $M = 0$ ; $Re a (u, u) = 0$ for every $u$ , so $a$ is coercive with no $α > 0$ : a nonzero $u$ would give $α ∥ u ∥^{2} \leq Re a (u, u) = 0$ (Bounded, coercive and symmetric sesquilinear forms, Hilbert space).

[F2]

Since $F \neq 0$ there is $v_{0} \in H$ with $F (v_{0}) \neq 0$ ; also $H \neq {0}$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Real and imaginary parts, complex conjugation, and modulus).

[F3]

A solution of $a (u, v) = F (v)$ for all $v \in H$ would in particular satisfy $a (u, v_{0}) = F (v_{0})$ (The Lax--Milgram theorem records the equation whose hypotheses fail here).

Proof

1.1F1

The form is bounded but not coercive: $∣ a (u, v) ∣ = 0 \leq 0 \cdot ∥ u ∥ ∥ v ∥$ shows the bound $M = 0$ , while for every $u \neq 0$ and every $α > 0$ one has $Re a (u, u) = 0 < α ∥ u ∥^{2}$ .

1.2F2

A datum with nonzero value: $F \neq 0$ means $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣ > 0$ , so some $v_{0}$ has $F (v_{0}) \neq 0$ .

2.1F1F3step 1.2∎

No solution: if $u \in H$ satisfied $a (u, v) = F (v)$ for all $v$ , then $0 = a (u, v_{0}) = F (v_{0}) \neq 0$ , a contradiction. Hence the equation has no solution, so neither existence nor uniqueness follows from boundedness alone; the estimate $∥ u ∥ \leq ∥ F ∥ / α$ of The Lax--Milgram theorem has no content without $α > 0$ , and the coercivity hypothesis there cannot be dropped.

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A coercive form need not be symmetric

Statement refuted

Assume Countable Choice for the Lax--Milgram conclusion. On $H = C^{2}$ with the standard inner product define $a (u, v) : = u_{1} \overline{v_{1}} + u_{2} \overline{v_{2}} + i u_{1} \overline{v_{2}} .$ Then $a$ is sesquilinear in the convention of Bounded, coercive and symmetric sesquilinear forms, bounded with $∣ a (u, v) ∣ \leq 2 ∥ u ∥ ∥ v ∥$ and coercive with $Re a (u, u) \geq \frac{1}{2} ∥ u ∥^{2}$ , so $α = \frac{1}{2}$ . It is not symmetric: with $e_{1} = (1, 0)$ , $e_{2} = (0, 1)$ one has $a (e_{1}, e_{2}) = i$ while $\overline{a (e_{2}, e_{1})} = 0$ . Hence the Lax--Milgram theorem The Lax--Milgram theorem applies to this nonsymmetric form, and symmetry is not needed for existence and uniqueness; the energy-minimisation corollary Symmetric Lax--Milgram is energy minimisation is the part that genuinely uses symmetry. The example is consistent with the abstract forcing remark Nonsymmetric Lax--Milgram is not a scalar minimisation principle.

Facts & Assumptions

Given: Countable Choice; the Hilbert space $H = C^{2}$ with the standard inner product $∥ u ∥^{2} = ∣ u_{1} ∣^{2} + ∣ u_{2} ∣^{2}$ ; the form $a (u, v) = u_{1} \overline{v_{1}} + u_{2} \overline{v_{2}} + i u_{1} \overline{v_{2}}$ ; and the vectors $e_{1} = (1, 0)$ , $e_{2} = (0, 1)$ .

[F1]

Sesquilinearity in the convention linear in the first argument and conjugate-linear in the second, with boundedness and coercivity as in Bounded, coercive and symmetric sesquilinear forms (Real and complex inner-product spaces and their induced length, Hilbert space).

[F2]

Scalar facts: $∣ z \overline{w} ∣ = ∣ z ∣ ∣ w ∣$ , $\overline{i} = - i$ , $i^{2} = - 1$ ; and for vectors in $C^{2}$ , $∣ z_{1} w_{1} ∣ + ∣ z_{2} w_{2} ∣ \leq (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})^{1 / 2} (∣ w_{1} ∣^{2} + ∣ w_{2} ∣^{2})^{1 / 2}$ by Cauchy--Schwarz (Real and imaginary parts, complex conjugation, and modulus, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F3]

Lax--Milgram applies to bounded coercive forms and does not assume symmetry; the energy-minimisation corollary does assume it (The Lax--Milgram theorem, Symmetric Lax--Milgram is energy minimisation, Nonsymmetric Lax--Milgram is not a scalar minimisation principle).

Proof

1.1F1F2

Sesquilinearity, boundedness and coercivity: for scalars $λ$ , $a (λ u, v) = λ a (u, v)$ and $a (u, λ v) = \overline{λ} a (u, v)$ directly from the definition, so $a$ is sesquilinear in the stated convention. Moreover $∣ a (u, v) ∣ \leq ∣ u_{1} ∣ ∣ v_{1} ∣ + ∣ u_{2} ∣ ∣ v_{2} ∣ + ∣ u_{1} ∣ ∣ v_{2} ∣$ , and Cauchy--Schwarz applied to the pairs $(∣ u_{1} ∣, ∣ u_{2} ∣)$ , $(∣ v_{1} ∣, ∣ v_{2} ∣)$ gives $∣ a (u, v) ∣ \leq (∣ u_{1} ∣ + ∣ u_{2} ∣) (∣ v_{1} ∣ + ∣ v_{2} ∣) \leq 2 ∥ u ∥ ∥ v ∥$ ; and $a (u, u) = ∣ u_{1} ∣^{2} + ∣ u_{2} ∣^{2} + i u_{1} \overline{u_{2}}$ has real part at least $∥ u ∥^{2} - ∣ u_{1} ∣ ∣ u_{2} ∣ \geq \frac{1}{2} ∥ u ∥^{2}$ , using $2 ∣ u_{1} ∣ ∣ u_{2} ∣ \leq ∣ u_{1} ∣^{2} + ∣ u_{2} ∣^{2}$ . Thus $a$ is coercive with constant $α = \frac{1}{2}$ .

1.2F1F2

Nonsymmetry: for the standard basis vectors, $a (e_{1}, e_{2}) = i$ while $\overline{a (e_{2}, e_{1})} = \overline{0} = 0$ , so $a (e_{1}, e_{2}) \neq \overline{a (e_{2}, e_{1})}$ and $a$ is not symmetric.

2.1F1F3step 1.1step 1.2∎

Consequences: $a$ is bounded and coercive but not symmetric, so Lax--Milgram applies and gives existence and uniqueness of solutions for every bounded conjugate-linear datum, while the energy-minimisation characterisation, which requires symmetry, does not apply. Thus symmetry is not needed for solvability but is genuinely used by the variational principle.

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Arbitrary $L^{2}$ boundary data need not have an $H^{1}$ lifting

Statement refuted

Assume the Axiom of Choice. Let $Ω \subset R^{2}$ be a bounded $C^{1}$ domain, let a boundary chart contain the closed straight segment $[0, 1]$ strictly inside its patch, and let $g : = 1_{(0, 1)}$ on that segment, extended by zero. Then $g \in L^{2} (\partial Ω)$ , but $g \notin H^{1 / 2} (\partial Ω) = W^{1 / 2, 2} (\partial Ω)$ : the Slobodeckij seminorm of the line jump at exponent $θ = 1 - 1 / 2$ diverges logarithmically, and by the sharp trace theorem the trace range of $W^{1, 2} (Ω)$ is exactly $W^{1 / 2, 2} (\partial Ω)$ . Consequently no $u \in H^{1} (Ω)$ has $T u = g$ , so the boundary-value problem with this $L^{2}$ datum is not solvable in $H^{1}$ : the lifting hypothesis of the weak Dirichlet formulation cannot be relaxed to arbitrary $L^{2}$ boundary data. No claim is made about the range $1 < p < 2$ , where the same jump function does lie in the trace space.

Facts & Assumptions

Given: The Axiom of Choice together with Countable Choice; a bounded $C^{1}$ domain $Ω \subset R^{2}$ with a boundary chart containing the closed straight segment $[0, 1]$ strictly inside its patch; the jump function $g = 1_{(0, 1)}$ on that segment, extended by zero; the surface measure on $\partial Ω$ ; and the exponent $θ = 1 - 1 / 2 = 1 / 2$ , so that $q : = p θ = 1$ at $p = 2$ . (Bounded C^k domains and boundary charts, Surface integration on compact C1 hypersurfaces, The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The boundary norm of The fractional Sobolev space on a compact $C^{1}$ boundary is a sum over a finite boundary atlas of the Euclidean Slobodeckij norms of the localised representations $(χ_{j} g) \circ Ψ_{j}^{- 1}$ , where $χ_{j}$ is a subordinate finite ambient partition; the Euclidean norm is that of The Gagliardo--Slobodeckij space on Euclidean space, the sum of the $L^{p}$ norm and the extended seminorm $[h]_{s, p} = (\int \int ∣ h (x) - h (y) ∣^{p} ∣ x - y ∣^{- 1 - s p} d x d y)^{1 / p}$ , and the set $W^{s, p} (\partial Ω)$ and its topology are independent of the atlas (Chart independence of the fractional boundary norm).

[F2]

Assume Countable Choice. For nonnegative measurable functions on a product of sigma-finite measure spaces the double integral equals the iterated integrals. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F3]

A bounded measurable function supported in a set of finite surface measure is an $L^{p} (\partial Ω)$ class for every $p$ . (The space $L^{p} (μ)$ as the quotient by null functions, Surface integration on compact C1 hypersurfaces)

[F4]

Sharp trace theorem: for $1 < p < \infty$ the trace operator $T : W^{1, p} (Ω) \to L^{p} (\partial Ω)$ of The $L^{p}$ trace operator on a bounded $C^{1}$ domain has range exactly $W^{1 - 1 / p, p} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact $C^{1}$ boundary).

Proof

1.1F3given

The datum is an $L^{2}$ class: the indicator of the straight segment $(0, 1)$ is bounded and is supported in a set of finite surface measure, so by [F3] it is an $L^{p} (\partial Ω)$ class for every $p$ , in particular $g \in L^{2} (\partial Ω)$ . At $p = 2$ the exponent is $θ = 1 / 2$ and $q = p θ = 1$ .

1.2F1F2algebra

The line-jump seminorm: for $h = 1_{(0, 1)}$ on $R$ and $q = p θ > 0$ the integrand $∣ h (x) - h (y) ∣^{p}$ is nonzero exactly when one of $x, y$ lies in $(0, 1)$ and the other does not. By symmetry and [F2] the double integral equals twice its part with $x \in (0, 1)$ , $y \notin (0, 1)$ , and the elementary antiderivative $\int u^{- 1 - q} d u = - u^{- q} / q$ gives $\int_{- \infty}^{0} (x - y)^{- 1 - q} d y + \int_{1}^{\infty} (y - x)^{- 1 - q} d y = \frac{x^{- q} + (1 - x)^{- q}}{q}$ for $x \in (0, 1)$ . Hence $[h]_{θ, p}^{p} = \frac{2}{q} \int_{0}^{1} (x^{- q} + (1 - x)^{- q}) d x = \frac{4}{q} \int_{0}^{1} x^{- q} d x,$ which is finite exactly when $q < 1$ ; translating and scaling the interval $(0, 1)$ to another interval $(a, b)$ changes this value by the finite factor $(b - a)^{1 - q}$ , so finiteness is intrinsic to the interval indicator.

2.1step 1.2algebra

Divergence at $p = 2$ : at $q = 1$ the integral in step 1.2 is $\int_{0}^{1} x^{- 1} d x = + \infty$ , diverging logarithmically at the endpoint $x = 0$ , so $[h]_{1 / 2, 2} = + \infty$ .

3.1F1step 2.1

The boundary norm is infinite: fix the finite atlas of [F1] so that it contains the given straight chart with a cutoff equal to one on the closed segment — possible because the segment lies strictly inside the patch — so that this chart's localised representation is the interval indicator $h$ of step 1.2. Then the corresponding summand of the boundary norm is $+ \infty$ while every other summand is nonnegative, so $∥ g ∥_{W^{1 / 2, 2} (\partial Ω)} = + \infty$ ; by the atlas independence in [F1] the space $W^{1 / 2, 2} (\partial Ω)$ is the same set for every atlas, so $g \notin W^{1 / 2, 2} (\partial Ω) = H^{1 / 2} (\partial Ω)$ .

4.1F4step 3.1∎

No $H^{1}$ lifting: at $p = 2$ the sharp trace theorem identifies the range of $T : W^{1, 2} (Ω) = H^{1} (Ω) \to L^{2} (\partial Ω)$ with $W^{1 / 2, 2} (\partial Ω)$ ; since $g$ lies outside this range, no $u \in H^{1} (Ω)$ satisfies $T u = g$ , and the inhomogeneous problem with this $L^{2}$ datum is not solvable in $H^{1}$ .

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The Neumann Poisson problem is not coercive on all of $H^{1}$

Statement refuted

Assume the Axiom of Choice inherited through the cited suppliers, together with Countable Choice. Let $Ω \subseteq R^{n}$ be a nonempty bounded open set and consider the form $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H^{1} (Ω)$ with the inner product of The Sobolev space $H^{1}$ is a Hilbert space. The constant function $1$ satisfies $a (1, 1) = 0$ while $∥ 1 ∥_{H^{1}} = ∣ Ω ∣^{1 / 2} > 0$ , so no $α > 0$ can satisfy $Re a (u, u) \geq α ∥ u ∥_{H^{1}}^{2}$ for all $u \in H^{1} (Ω)$ : the form is not coercive on $H^{1} (Ω)$ , and Lax--Milgram does not apply in that space. The obstruction is exactly the kernel: $a (u, u) = 0$ forces $\nabla u = 0$ , hence $u$ is constant on each connected component (Zero weak gradient gives componentwise constants), and the associated Neumann problem $a (u, v) = F (v)$ for all $v$ can have no solution when $F (1) \neq 0$ while constants give nontrivial solutions of the homogeneous equation. This motivates the mean-zero subspace formulation Weak Neumann solvability on the mean-zero subspace and its compatibility condition.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the Hilbert space $H^{1} (Ω)$ with inner product $(u, v)_{H^{1}} = (u, v)_{L^{2}} + \sum_{i} (D_{i} u, D_{i} v)_{L^{2}}$ ; the form $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ ; and the constant class $1$ .

[F1]

Coercivity of a sesquilinear form means $Re a (u, u) \geq α ∥ u ∥_{H^{1}}^{2}$ for all $u$ and some $α > 0$ ; boundedness means the same form has a finite bound (Bounded, coercive and symmetric sesquilinear forms).

[F2]

$1 \in H^{1} (Ω)$ with weak gradient $0$ : the classical partial derivatives of the constant are $0$ and are its weak derivatives, and the constant is in $L^{2}$ because $Ω$ has finite measure (Classical derivatives agree with weak derivatives, Euclidean balls have positive finite Lebesgue measure, Integer-order Sobolev spaces and their norms).

[F3]

$∥ 1 ∥_{H^{1}}^{2} = ∥ 1 ∥_{L^{2}}^{2} + ∥ D 1 ∥_{L^{2}}^{2} = ∣ Ω ∣ + 0 > 0$ , since $∣ Ω ∣ > 0$ for a nonempty open set (Euclidean balls have positive finite Lebesgue measure, Integral over a measurable subset, Hilbert space).

[F4]

Zero weak gradient implies componentwise constancy: $a (u, u) = 0$ means $\int_{Ω} ∣ \nabla u ∣^{2} = 0$ , so $\nabla u = 0$ a.e. and $u$ is constant on each connected component of $Ω$ (Zero weak gradient gives componentwise constants, Connected components, quasicomponents, and totally disconnected spaces, Real and imaginary parts, complex conjugation, and modulus).

[F5]

The mean-zero Neumann theorem requires the compatibility $F (1) = 0$ and produces solutions with $\int_{Ω} u = 0$ (Weak Neumann solvability on the mean-zero subspace).

Proof

1.1F2F3

The constant is not infinitesimal for the form: by [F2] the weak gradient of $1$ vanishes, so $a (1, 1) = \int_{Ω} ∣ \nabla 1 ∣^{2} d x = 0$ , while $∥ 1 ∥_{H^{1}}^{2} = ∣ Ω ∣ > 0$ by [F3].

2.1F1step 1.1

Failure of coercivity: if some $α > 0$ satisfied $Re a (u, u) \geq α ∥ u ∥_{H^{1}}^{2}$ for all $u$ , then at $u = 1$ it would give $0 = Re a (1, 1) \geq α ∣ Ω ∣ > 0$ , a contradiction. Hence the form is not coercive on $H^{1} (Ω)$ , and the Lax--Milgram existence theorem does not apply in that space.

3.1F4F5step 2.1∎

The obstruction is the kernel and the compatibility: by [F4], $a (u, u) = 0$ forces $\nabla u = 0$ a.e., so with $u \neq 0$ the constants are nontrivial solutions of the homogeneous equation; the weak equation $a (u, v) = F (v)$ on all of $H^{1} (Ω)$ , tested at $v = 1$ , forces $F (1) = 0$ , so no solution exists when $F (1) \neq 0$ . On a connected extension domain this obstruction is removed by the cited mean-zero formulation and compatibility condition. On a disconnected domain one must remove constants on every component and impose compatibility on each component; global mean zero alone does not remove the kernel.

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Complex sesquilinear coercivity differs from bilinear positivity

Example

On $H = C$ define $a (u, v) : = u \overline{v}$ and $b (u, v) : = u v$ . Then $a$ is sesquilinear in the convention of Bounded, coercive and symmetric sesquilinear forms (linear in the first argument, conjugate-linear in the second), bounded with $M = 1$ , and coercive with constant $α = 1$ because $a (u, u) = ∣ u ∣^{2}$ . The expression $b$ is bilinear, not conjugate-linear in the second argument, and it fails the coercivity condition: $b (i, i) = - 1$ , so $Re b (u, u) \geq α ∣ u ∣^{2}$ fails at $u = i$ for every $α > 0$ . More generally $b (u, u) = u^{2}$ is not real for $u \notin R \cup i R$ , and its real part is negative, for example, at $u = 1 + 2 i$ , since $(1 + 2 i)^{2} = - 3 + 4 i$ . Hence the conjugation in the second slot is not cosmetic: the complex Lax--Milgram hypotheses cannot be applied to this bilinear pairing $b$ , and the real bilinear convention of Bounded, coercive and symmetric sesquilinear forms is genuinely a different hypothesis. This tests exactly the convention on which The Lax--Milgram theorem and A bounded form is represented by a unique bounded operator depend, and complements the real-form sources [Si] and [H], which state real bilinear versions.

Facts & Assumptions

Given: The Hilbert space $H = C$ with its usual inner product and $∣ z ∣$ the complex modulus; the pairings $a (u, v) = u \overline{v}$ and $b (u, v) = u v$ .

[F1]

Sesquilinearity, boundedness and coercivity definitions: $a$ is linear in the first slot and conjugate-linear in the second with $∣ a (u, v) ∣ \leq M ∣ u ∣ ∣ v ∣$ , and coercive with constant $α$ when $Re a (u, u) \geq α ∣ u ∣^{2}$ (Bounded, coercive and symmetric sesquilinear forms, Real and complex inner-product spaces and their induced length, Hilbert space).

[F2]

Scalar facts: $u \overline{u} = ∣ u ∣^{2} \geq 0$ , $∣ u \overline{v} ∣ = ∣ u ∣ ∣ v ∣$ , $\overline{i} = - i$ and $i^{2} = - 1$ (Real and imaginary parts, complex conjugation, and modulus).

[F3]

Lax--Milgram and the form-to-operator lemma are stated for sesquilinear forms in the conjugate-linear-second-slot convention (The Lax--Milgram theorem, A bounded form is represented by a unique bounded operator).

Proof

1.1F1F2

The sesquilinear form $a$ : for scalars $λ$ , $a (λ u, v) = λ u \overline{v} = λ a (u, v)$ and $a (u, λ v) = u \overline{λ v} = \overline{λ} a (u, v)$ , so $a$ is linear in the first argument and conjugate-linear in the second; $∣ a (u, v) ∣ = ∣ u ∣ ∣ v ∣ \leq 1 \cdot ∣ u ∣ ∣ v ∣$ gives the bound $M = 1$ , and $a (u, u) = ∣ u ∣^{2}$ gives coercivity with $α = 1$ .

1.2F2F1

The bilinear pairing $b$ is not of this type: $b (λ u, v) = λ b (u, v) = b (u, λ v)$ , so $b$ is bilinear; but with $u = v = 1$ and $λ = i$ , $b (1, i) = i \neq - i = \overline{i} b (1, 1)$ , so $b$ is not conjugate-linear in the second slot.

2.1F2step 1.2algebra

$b$ fails coercivity: $b (i, i) = i^{2} = - 1$ has real part $- 1$ , while $α ∣ i ∣^{2} = α > 0$ for every $α > 0$ ; hence $Re b (u, u) \geq α ∣ u ∣^{2}$ fails at $u = i$ for every $α$ . More generally $b (u, u) = u^{2}$ is not real unless $u \in R \cup i R$ .

3.1F1F2F3step 1.2step 2.1algebra∎

Consequences: neither Lax--Milgram nor the representation lemma applies to this $b$ , since it is not sesquilinear. More generally, a complex form that is both bilinear and sesquilinear satisfies $i c (u, v) = c (u, i v) = - i c (u, v)$ , hence is the zero form. The zero form satisfies the bounded sesquilinear hypotheses of the representation lemma; on a nonzero space it cannot be coercive, but on $H = {0}$ it is coercive with every $α > 0$ and satisfies the Lax--Milgram form hypotheses. Thus the conjugation convention matters, with this zero-form exception.

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A one-dimensional form attains the $1 / α$ Lax--Milgram bound

Example

On $H = C$ with the standard inner product and $α > 0$ , let $a (u, v) : = α u \overline{v}$ and let $F_{c} (v) : = c \overline{v}$ for a fixed $c \in C$ . Then $a$ is bounded with $M = α$ , coercive with the same constant $α$ , and the Lax--Milgram solution of $a (u, v) = F_{c} (v)$ for all $v$ is $u = \frac{c}{α},$ since $α u \overline{v} = c \overline{v}$ for all $v$ forces $α u = c$ . The solution operator has norm exactly $1 / α$ : $∥ F_{c} ∥ = \sup_{∣ v ∣ \leq 1} ∣ c \overline{v} ∣ = ∣ c ∣$ and $∣ u ∣ = ∣ c ∣ / α$ , so $∥ S ∥ = \frac{1}{α} .$ Hence the bound of The Lax--Milgram solution operator has norm at most $1 / α$ is attained and cannot be improved uniformly over coercive forms; this is the plan’s sharpness example and the one-dimensional model of the general estimate.

Facts & Assumptions

Given: A real $α > 0$ ; the Hilbert space $H = C$ with its usual inner product and modulus; the form $a (u, v) = α u \overline{v}$ ; and the functional $F_{c} (v) = c \overline{v}$ for a fixed $c \in C$ .

[F1]

$a$ is sesquilinear, bounded with $M = α$ , and coercive with the same constant: $∣ a (u, v) ∣ = α ∣ u ∣ ∣ v ∣$ and $a (u, u) = α ∣ u ∣^{2}$ (Bounded, coercive and symmetric sesquilinear forms, Real and imaginary parts, complex conjugation, and modulus, Hilbert space).

[F2]

Every conjugate-linear functional on $C$ has the form $F_{c} (v) = \overline{v} F_{c} (1)$ ; testing at $v = 1$ directly determines the unique solution. The abstract comparison is The Lax--Milgram solution operator has norm at most $1 / α$ , but no choice principle is needed for this scalar computation.

[F3]

Operator norm: $∥ F_{c} ∥ = \sup_{∣ v ∣ \leq 1} ∣ c \overline{v} ∣ = ∣ c ∣$ and $∥ S ∥ = \sup {∥ S (F) ∥ : ∥ F ∥ \leq 1}$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

1.1F1F2algebra

Direct solution: the equation $a (u, v) = F_{c} (v)$ reads $α u \overline{v} = c \overline{v}$ for every $v \in C$ . Testing with $v = 1$ forces $α u = c$ , that is $u = c / α$ ; conversely this $u$ satisfies the equation for every $v$ . The scalar equation also proves uniqueness directly.

1.2F1

Boundedness and coercivity constants: from $∣ a (u, v) ∣ = α ∣ u ∣ ∣ v ∣$ the least bound is $M = α$ , and coercivity holds with $α$ since $a (u, u) = α ∣ u ∣^{2}$ ; no larger coercivity constant can work at $u = 1$ .

2.1F2F3step 1.1

Norms: $∥ F_{c} ∥ = ∣ c ∣$ by [F3], and $S (F_{c}) = c / α$ has modulus $∣ c ∣ / α$ , so $∥ S (F_{c}) ∥ / ∥ F_{c} ∥ = 1 / α$ for every $c \neq 0$ ; hence $∥ S ∥ = 1 / α$ , attaining the bound of the corollary.

3.1step 1.2step 2.1∎

Conclusion: the estimate $∥ S ∥ \leq 1 / α$ is sharp and cannot be improved uniformly over bounded coercive forms on a fixed Hilbert space; the one-dimensional computation is the model of the general constant.

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The Neumann kernel is spanned by the componentwise constants

Example

Assume the Axiom of Choice inherited through the cited suppliers, together with Countable Choice. Let $Ω \subseteq R^{n}$ be a nonempty bounded extension domain (Sobolev extension domains and extension operators) whose connected components are $Ω_{1}, \dots, Ω_{m}$ , and let $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H^{1} (Ω)$ . Then ${u \in H^{1} (Ω) : a (u, v) = 0 for all v \in H^{1} (Ω)} = {u \in H^{1} (Ω) : \nabla u = 0 a.e.} = {span}_{K} {1_{Ω_{1}}, \dots, 1_{Ω_{m}}},$ the space of classes constant on each connected component. The indicators are linearly independent because they are nonzero on disjoint sets of positive measure, so the kernel is $m$ -dimensional; for a connected $Ω$ it is exactly the constants and the Neumann form has a one-dimensional kernel. This refines the connected-domain constants warning on the base page and motivates the per-component compatibility condition recorded in Weak Neumann solvability on the mean-zero subspace; the plan's B-page example states it so that no separate dimension theory is needed.

Facts & Assumptions

Given: The Axiom of Choice; a nonempty bounded $W^{1, 2}$ -extension domain $Ω \subseteq R^{n}$ , $n \geq 1$ , with connected components $Ω_{1}, \dots, Ω_{m}$ ( $m \geq 1$ ); the form $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H^{1} (Ω) = W^{1, 2} (Ω; K)$ , where $\nabla u = (D_{1} u, \dots, D_{n} u)$ (Integer-order Sobolev spaces and their norms, Sobolev extension domains and extension operators, Integral over a measurable subset, The Axiom of Choice).

[F1]

The Axiom of Choice supplies Countable Choice, the interface used by the Sobolev and Lebesgue suppliers below (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Testing and nonnegativity: $\int_{Ω} ∣ \nabla u ∣^{2} d x = \sum_{j = 1}^{n} \int_{Ω} ∣ D_{j} u ∣^{2} d x \geq 0$ , and a nonnegative measurable integral vanishes exactly when its integrand vanishes almost everywhere; on the a.e. quotient a bounded function is an $L^{2}$ class when the underlying set has finite measure (Integral over a measurable subset, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

Zero weak gradient implies componentwise constancy: if $u \in W^{1, 2} (Ω; K)$ has $D_{j} u = 0$ a.e. for every $j$ , then for every connected component $C$ of $Ω$ there is $c_{C} \in K$ with $u = c_{C}$ a.e. on $C$ (Zero weak gradient gives componentwise constants).

[F4]

Components and geometry: every connected component of an open Euclidean set is open and connected, and a nonempty open set contains a Euclidean ball; every Euclidean ball has positive finite Lebesgue measure (Every connected component of an open subset of $R^{n}$ is open and polygonally connected, Connected components, quasicomponents, and totally disconnected spaces, Euclidean balls have positive finite Lebesgue measure).

[F5]

Classical derivatives are weak derivatives: a function whose real and imaginary parts are of class $C^{k}$ has its classical partial derivatives of order $\leq k$ as weak derivatives, and the classical partial derivative at a point of a locally constant function vanishes (the difference quotients are eventually zero) (Classical derivatives agree with weak derivatives, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Proof

1.1F1F2

The kernel is the zero-gradient set. Let $u \in H^{1} (Ω)$ satisfy $a (u, v) = 0$ for every $v \in H^{1} (Ω)$ . Testing with $v = u$ gives $0 = a (u, u) = \int_{Ω} ∣ \nabla u ∣^{2} d x = \sum_{j = 1}^{n} \int_{Ω} ∣ D_{j} u ∣^{2} d x$ , a finite sum of nonnegative terms, so each $\int_{Ω} ∣ D_{j} u ∣^{2} d x = 0$ and hence $D_{j} u = 0$ almost everywhere for every $j$ . Conversely, if $D_{j} u = 0$ a.e. for all $j$ then $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x = 0$ for every $v \in H^{1} (Ω)$ , because a function vanishing a.e. has zero integral against every class. So the kernel equals ${u \in H^{1} (Ω) : \nabla u = 0 a.e.}$ .

2.1F3F4F5step 1.1algebra

Identification with the componentwise constants. Let $u \in H^{1} (Ω)$ have $\nabla u = 0$ a.e. Since $u \in W^{1, 2} (Ω; K)$ and all weak first derivatives vanish a.e., the componentwise constancy theorem gives, for each component $Ω_{i}$ , a constant $c_{i} \in K$ with $u = c_{i}$ almost everywhere on $Ω_{i}$ ; as the components partition $Ω$ , $u = \sum_{i = 1}^{m} c_{i} 1_{Ω_{i}}$ almost everywhere. Conversely let $c_{1}, \dots, c_{m} \in K$ and put $w : = \sum_{i = 1}^{m} c_{i} 1_{Ω_{i}}$ on $Ω$ . Each component is open, so every point $x \in Ω$ has the open neighbourhood $Ω_{i (x)}$ on which $w$ is constant; hence all classical partial derivatives of $w$ exist at every point of $Ω$ and vanish, and they are the weak derivatives by [F5]. Moreover $w$ is bounded and $Ω$ is bounded, hence has finite measure, so $w$ is an $L^{2}$ class; therefore $w \in H^{1} (Ω)$ with $D_{j} w = 0$ a.e. for every $j$ , and step 1.1 puts $w$ in the kernel.

3.1F4step 2.1algebra

Independence and dimension. Each component $Ω_{i}$ is nonempty, hence contains a Euclidean ball of positive measure, and $w = \sum_{i} c_{i} 1_{Ω_{i}}$ equals the constant $c_{i}$ everywhere on $Ω_{i}$ . If $w = 0$ as an $L^{2}$ class and $c_{i} \neq 0$ for some $i$ , then $w$ would be nonzero on the positive-measure set $Ω_{i}$ while the zero class vanishes almost everywhere, a contradiction; hence every $c_{i} = 0$ . So the $m$ indicators are linearly independent, the space ${u : \nabla u = 0 a.e.}$ is exactly their span, and the kernel of the Neumann form is $m$ -dimensional; for connected $Ω$ ( $m = 1$ ) it is the one-dimensional space of constants.

4.1step 1.1step 2.1step 3.1∎

Conclusion: the kernel of $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $H^{1} (Ω)$ is the $m$ -dimensional space of classes constant on each connected component, motivating the per-component compatibility condition for the Neumann problem; no dimension theory beyond this display is used.

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A large adverse zero-order term destroys Dirichlet coercivity

Statement refuted

Assume the Axiom of Choice inherited through the cited general solvability theorem, together with Countable Choice. Let $I = (0, 1)$ , $c \geq 0$ and $a_{c} (u, v) = \int_{0}^{1} u^{'} \overline{v^{'}} d x - c \int_{0}^{1} u \overline{v} d x (u, v \in H_{0}^{1} (I)) .$ The sharp constant is $C_{P} = 1 / π$ by The sharp Dirichlet Poincare inequality on an interval. For $0 \leq c < π^{2}$ , the form is coercive with constant $(π^{2} - c) / (1 + π^{2})$ in the standard $H^{1}$ norm, and The Lax--Milgram theorem gives a unique weak solution for every bounded conjugate-linear functional. For $c \geq π^{2}$ , the nonzero test $ϕ (x) = \sin (π x)$ gives $a_{c} (ϕ, ϕ) = (π^{2} - c) ∥ ϕ ∥_{2}^{2} \leq 0$ , so the form is not coercive. At the endpoint $c = π^{2}$ , the helper's weak identity gives $a_{c} (ϕ, v) = 0$ for every $v \in H_{0}^{1} (I)$ : both $0$ and $ϕ$ solve the homogeneous weak Dirichlet problem. The original polynomial witness also remains valid: $p (x) = x (1 - x) \in H_{0}^{1} (I)$ satisfies $∥ p^{'} ∥_{2}^{2} = 1 / 3$ and $∥ p ∥_{2}^{2} = 1 / 30$ , hence $a_{c} (p, p) = 1 / 3 - c / 30 \leq 0$ for $c \geq 10$ . Thus the lower-order sign/smallness mechanism in Lax--Milgram solvability for coercive divergence-form equations cannot be omitted. In this interval model its energy argument with the local sharp constant gives the exact coercivity condition $c < 1 / C_{P}^{2} = π^{2}$ ; the generic Poincare supplier itself is not claimed to provide that numerical constant.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the interval $I = (0, 1)$ ; a real constant $c \geq 0$ ; the form $a_{c} (u, v) = \int_{0}^{1} u^{'} \overline{v^{'}} d x - c \int_{0}^{1} u \overline{v} d x$ on $H_{0}^{1} (I)$ ; and the helper function $ϕ (x) = \sin (π x)$ .

[F1]

Sharp interval inequality and witness: $∥ u ∥_{L^{2}} \leq (1 / π) ∥ u^{'} ∥_{L^{2}}$ for $u \in H_{0}^{1} (I)$ ; $ϕ \in H_{0}^{1} (I)$ is nonzero, $ϕ^{'} ϕ$ -identities hold, and $\int_{0}^{1} ϕ^{'} \overline{v^{'}} d x = π^{2} \int_{0}^{1} ϕ \overline{v} d x$ for every $v \in H_{0}^{1} (I)$ (The sharp Dirichlet Poincare inequality on an interval).

[F2]

Hilbert structure: $H_{0}^{1} (I)$ is a Hilbert space with $∥ u ∥_{H^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ u^{'} ∥_{L^{2}}^{2}$ , and bounded coercive forms on it have unique solutions for every bounded conjugate-linear datum (The Sobolev space $H^{1}$ is a Hilbert space, The Lax--Milgram theorem, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

Estimates: $∣ a_{c} (u, v) ∣ \leq ∥ u^{'} ∥_{L^{2}} ∥ v^{'} ∥_{L^{2}} + c ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq (1 + c) ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ by H"older; and $∥ u ∥_{H^{1}}^{2} \leq (1 + 1 / π^{2}) ∥ u^{'} ∥_{L^{2}}^{2}$ by [F1] (Holder's inequality for integrals, including the endpoint cases, Complex Holder, Minkowski, and the quotient norm, Bounded, coercive and symmetric sesquilinear forms).

[F4]

Cutoff construction on the interval: the standard smooth step $σ$ has $σ^{'} \equiv 0$ outside $(0, 1)$ and $C_{σ} : = \sup ∣ σ^{'} ∣ < \infty$ ; chain and product rules give derivatives of $η_{m} (x) = σ (m x - 1) σ (m (1 - x) - 1)$ ; elementary interval bounds, additivity over subintervals, linearity of the integral, and the agreement of the Riemann and Lebesgue integrals for bounded Riemann integrable functions on a closed interval control the resulting $L^{2}$ norms (The standard smooth step function, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value, If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[F5]

The fundamental theorem of calculus and the absolutely continuous representative of a one-dimensional Sobolev class control the polynomial integrals below (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives, Integral over a measurable subset, The space $L^{p} (μ)$ as the quotient by null functions).

Proof

1.1F1F2F3algebra

Coercivity below the threshold: for $0 \leq c < π^{2}$ and $u \in H_{0}^{1} (I)$ , $a_{c} (u, u) = ∥ u^{'} ∥_{L^{2}}^{2} - c ∥ u ∥_{L^{2}}^{2} \geq (1 - c / π^{2}) ∥ u^{'} ∥_{L^{2}}^{2}$ by [F1], and $∥ u ∥_{H^{1}}^{2} \leq (1 + 1 / π^{2}) ∥ u^{'} ∥_{L^{2}}^{2}$ , so $a_{c} (u, u) \geq \frac{π^{2} - c}{1 + π^{2}} ∥ u ∥_{H^{1}}^{2}$ : the form is coercive with constant $(π^{2} - c) / (1 + π^{2})$ and bounded by [F3]; Lax--Milgram gives a unique solution for every bounded conjugate-linear functional.

1.2F1algebra

Failure at and above the threshold: the helper witness satisfies $ϕ \in H_{0}^{1} (I)$ , $ϕ \neq 0$ , and $a_{c} (ϕ, ϕ) = ∥ ϕ^{'} ∥_{L^{2}}^{2} - c ∥ ϕ ∥_{L^{2}}^{2} = (π^{2} - c) ∥ ϕ ∥_{L^{2}}^{2}$ by the weak identity of [F1] with $v = ϕ$ . For $c \geq π^{2}$ this is at most $0$ while $ϕ \neq 0$ , so no $α > 0$ can satisfy $Re a_{c} (u, u) \geq α ∥ u ∥^{2}$ for all $u$ : coercivity fails.

1.3F4F5algebra

Polynomial witness: let $p (x) = x (1 - x)$ . Then $p$ is smooth on $[0, 1]$ , $p (0) = p (1) = 0$ , $∣ p (x) ∣ \leq \min (x, 1 - x)$ and $∣ p^{'} (x) ∣ \leq 1$ ; for $m \geq 4$ put $η_{m} (x) = σ (m x - 1) σ (m (1 - x) - 1)$ and $p_{m} : = η_{m} p \in C_{c}^{\infty} (I)$ . As in [F4], $∣ η_{m}^{'} ∣ \leq 2 m C_{σ}$ , $p_{m} = p$ on $[2 / m, 1 - 2 / m]$ , and on the two endpoint strips $∣ p_{m} - p ∣ \leq 2 / m$ and $∣ (p_{m} - p)^{'} ∣ \leq 4 C_{σ} + 1$ ; hence $∥ p_{m} - p ∥_{L^{2}}^{2} \leq 4 / m^{2}$ and $∥ (p_{m} - p)^{'} ∥_{L^{2}}^{2} \leq (4 C_{σ} + 1)^{2} \cdot 4 / m$ , so $p_{m} \to p$ in $H^{1}$ and $p \in H_{0}^{1} (I)$ . The fundamental theorem and linearity give $\int_{0}^{1} p^{' 2} = \int_{0}^{1} (1 - 2 x)^{2} d x = 1 - 2 + \frac{4}{3} = \frac{1}{3}$ and $\int_{0}^{1} p^{2} = \int_{0}^{1} (x^{2} - 2 x^{3} + x^{4}) d x = \frac{1}{3} - \frac{1}{2} + \frac{1}{5} = \frac{1}{30}$ ; hence $a_{c} (p, p) = \frac{1}{3} - \frac{c}{30} \leq 0$ for $c \geq 10$ .

2.1F1step 1.2

Endpoint nonuniqueness: at $c = π^{2}$ the same weak identity gives $a_{π^{2}} (ϕ, v) = 0$ for every $v \in H_{0}^{1} (I)$ ; since $ϕ \neq 0$ , both the zero function and $ϕ$ solve the homogeneous weak Dirichlet problem, so uniqueness fails at the endpoint. No claim is made here about nonuniqueness for $c > π^{2}$ .

3.1step 1.1step 2.1step 1.3∎

Conclusion: for $0 \leq c < π^{2}$ the form is coercive with the explicit constant and Lax--Milgram applies; for $c \geq π^{2}$ the nonzero sine witness destroys coercivity with equality of the quadratic form on $ϕ$ at the endpoint, where nonuniqueness is explicit; the polynomial witness independently witnesses failure for $c \geq 10$ . Therefore the sign/smallness mechanism of the general solvability theorem cannot be omitted, and in this interval model the exact threshold is $c < 1 / C_{P}^{2} = π^{2}$ .

Lax--Milgram and Weak Elliptic Solutions — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The weak Dirichlet Poisson problem on an interval

Example

Facts & Assumptions

Proof

L2L2 forcing defines an H−1H−1 functional

Example

Facts & Assumptions

Proof

A nonsymmetric coercive elliptic form

Example

Facts & Assumptions

Proof

A bounded form without coercivity need not be solvable

Statement refuted

Facts & Assumptions

Proof

A coercive form need not be symmetric

Statement refuted

Facts & Assumptions

Proof

Arbitrary L2L2 boundary data need not have an H1H1 lifting

Statement refuted

Facts & Assumptions

Proof

The Neumann Poisson problem is not coercive on all of H1H1

Statement refuted

Facts & Assumptions

Proof

Complex sesquilinear coercivity differs from bilinear positivity

Example

Facts & Assumptions

Proof

A one-dimensional form attains the 1/α1/α Lax--Milgram bound

Example

Facts & Assumptions

Proof

The Neumann kernel is spanned by the componentwise constants

Example

Facts & Assumptions

Proof

A large adverse zero-order term destroys Dirichlet coercivity

Statement refuted

Facts & Assumptions

Proof

Sources

$L^{2}$ forcing defines an $H^{- 1}$ functional

Arbitrary $L^{2}$ boundary data need not have an $H^{1}$ lifting

The Neumann Poisson problem is not coercive on all of $H^{1}$

A one-dimensional form attains the $1 / α$ Lax--Milgram bound