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The Neumann Poisson problem is not coercive on all of H1

Statement refuted

Assume the Axiom of Choice inherited through the cited suppliers, together with Countable Choice. Let Ω⊆Rn be a nonempty bounded open set and consider the form a(u,v)=∫Ω∇u⋅∇v‾ dx on H1(Ω) with the inner product of The Sobolev space H1 is a Hilbert space. The constant function 1 satisfies a(1,1)=0 while ∥1∥H1=∣Ω∣1/2>0, so no α>0 can satisfy Re⁡a(u,u)≥α∥u∥H12 for all u∈H1(Ω): the form is not coercive on H1(Ω), and Lax--Milgram does not apply in that space. The obstruction is exactly the kernel: a(u,u)=0 forces ∇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)≠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⊆Rn; the Hilbert space H1(Ω) with inner product (u,v)H1=(u,v)L2+∑i(Diu,Div)L2; the form a(u,v)=∫Ω∇u⋅∇v‾ dx; and the constant class 1.

[F1]

Coercivity of a sesquilinear form means Re⁡a(u,u)≥α∥u∥H12 for all u and some α>0; boundedness means the same form has a finite bound (Bounded, coercive and symmetric sesquilinear forms).

[F2]

1∈H1(Ω) with weak gradient 0: the classical partial derivatives of the constant are 0 and are its weak derivatives, and the constant is in L2 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∥H12=∥1∥L22+∥D1∥L22=∣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 ∫Ω∣∇u∣2=0, so ∇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 ∫Ωu=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)=∫Ω∣∇1∣2 dx=0, while ∥1∥H12=∣Ω∣>0 by [F3].

2.1F1step 1.1

Failure of coercivity: if some α>0 satisfied Re⁡a(u,u)≥α∥u∥H12 for all u, then at u=1 it would give 0=Re⁡a(1,1)≥α∣Ω∣>0, a contradiction. Hence the form is not coercive on H1(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 ∇u=0 a.e., so with u≠0 the constants are nontrivial solutions of the homogeneous equation; the weak equation a(u,v)=F(v) on all of H1(Ω), tested at v=1, forces F(1)=0, so no solution exists when F(1)≠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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