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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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Sources

Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes) (standard reference, not scraped)
Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete 614-page text) (standard reference, not scraped)