Weak Neumann solvability on the mean-zero subspace

$V$ is a closed subspace: $V=\ker \mathrm{\ell }$ for the bounded linear functional $\mathrm{\ell }$ of [F2], hence closed; being a linear subspace of the Hilbert space $H^1(\mathrm{\Omega })$, it is itself a Hilbert space for the restricted inner product by [F7].

Coercivity on $V$: for $v\in V$, Poincar'e--Wirtinger gives $\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}^2=\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}^2+\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^2}^2\le (1+C_W^2)\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^2}^2$, so $\mathrm{Re}a(v,v)=\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^2}^2\ge \frac{1}{1+C_W^2}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}^2$; also $\mathrm{\mid }a(u,v)\mathrm{\mid }\le \mathrm{\parallel }Du{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$, so $a$ is bounded on $V$ with bound $1$.

Finiteness of components in the disconnected case. For a nonempty bounded $W^{1,2}$-extension domain, every component $C$ has positive measure by [F2] and its indicator belongs to $H^1$ with zero gradient by [F8]. If there were infinitely many components, AC would select distinct $C_j$, $j\ge 1$; the normalized indicators $e_j=\mathrm{\mid }{C}_j{\mathrm{\mid }}^{-1\mathrm{/}2}{\mathbf{1}}_{C_j}$ have $H^1$ norm $1$ and pairwise $L^2$ distance $\sqrt{2}$, contradicting [F8]. Hence the components are $C_1,\dots ,C_m$. On the closed subspace $V_c=\{v:{\int }_{C_k}v=0\text{ for every }k\}$ there is a constant $C_c$ with $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le C_c\mathrm{\parallel }Dv{\mathrm{\parallel }}_2$. Otherwise AC selects $v_j\in V_c$, $j\ge 1$, with $\mathrm{\parallel }{v}_j{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }D{v}_j{\mathrm{\parallel }}_2<1\mathrm{/}j$. By [F8] a subsequence converges in $L^2$; it is Cauchy in $H^1$, so [F1] gives an $H^1$ limit $v$ with $Dv=0$ and $\mathrm{\parallel }v{\mathrm{\parallel }}_2=1$. Each component integral passes to the limit by H"older, so $v\in V_c$; [F6] makes it constant on each component, hence zero by its componentwise mean, a contradiction.

Solution on $V$: the restriction $F{\mathrm{\mid }}_V$ is a bounded conjugate-linear functional on the Hilbert space $V$ and $a$ is bounded and coercive there with constant $\alpha :=1\mathrm{/}(1+C_W^2)$; Lax--Milgram gives a unique $u\in V$ with $a(u,v)=F(v)$ for every $v\in V$, satisfying $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le \mathrm{\parallel }F{\mathrm{\mid }}_V\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$.

Extension to all test functions: let $v\in H^1(\mathrm{\Omega })$ and put $c:=\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{-1}{\int }_{\mathrm{\Omega }}vdx$ and $v_0:=v-c\mathbf{1}$, so that $\int v_0=0$ and $v_0\in V$, while $\mathbf{1}\in H^1$ has zero weak gradient by [F4]. Then $a(u,v)=a(u,v_0)+a(u,c\mathbf{1})=a(u,v_0)$, and conjugate-linearity of $F$ together with $F(\mathbf{1})=0$ gives $F(v)=F(v_0)+\bar{c}F(\mathbf{1})=F(v_0)$; hence $a(u,v)=F(v)$ for every $v\in H^1(\mathrm{\Omega })$.

Uniqueness in $V$: if $u_1,u_2\in V$ both solve, then $w:=u_1-u_2\in V$ satisfies $a(w,v)=0$ for all $v\in V$; testing $v=w$ and using step 1.2 gives $\alpha \mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2\le \mathrm{Re}a(w,w)=0$, so $w=0$.

Estimate: from step 2.1, $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le (1+C_W^2)\mathrm{\parallel }F{\mathrm{\mid }}_V\mathrm{\parallel }\le (1+C_W^2)\mathrm{\parallel }F\mathrm{\parallel }$, the last inequality because the supremum over the smaller set $V$ is at most the supremum over $H^1(\mathrm{\Omega })$.

Full solution set and necessity: if $u^{\mathrm{\prime }}$ is any solution of $a(u^{\mathrm{\prime }},v)=F(v)$ on $H^1(\mathrm{\Omega })$, then $w:=u^{\mathrm{\prime }}-u$ satisfies $a(w,v)=0$ for all $v$; testing $v=w$ gives $\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^2}^2=0$, so $\mathrm{\nabla }w=0$ a.e. and, $\mathrm{\Omega }$ being connected, [F6] makes $w$ a constant class; hence the solution set is $u+\mathbb{K}\mathbf{1}$, and conversely every $u+c\mathbf{1}$ solves because $\mathbf{1}$ has zero weak gradient. Testing $v=\mathbf{1}$ in the equation gives $a(u,\mathbf{1})=0=F(\mathbf{1})$, so the compatibility $F(\mathbf{1})=0$ is necessary.

Componentwise solvability. The inequality of step 1.3 gives coercivity on $V_c$ with constant $1\mathrm{/}(1+C_c^2)$, so the argument of steps 1.1–2.1 gives a unique $u\in V_c$ solving there. Every $v\in H^1$ decomposes as $v=v_0+{\sum }_k{c}_k{\mathbf{1}}_{C_k}$, where $c_k=\mathrm{\mid }{C}_k{\mathrm{\mid }}^{-1}{\int }_{C_k}v$ and $v_0\in V_c$. Thus if $F({\mathbf{1}}_{C_k})=0$ for every $k$, the equation extends to all tests as in step 3.1; conversely testing each indicator makes these conditions necessary. Testing the difference of two solutions with itself and using [F6] shows that all solutions differ by componentwise constants, so zero mean on each component specifies the unique normalized solution.

This proves the connected-domain assertion and its displayed estimate, and establishes the stated componentwise compatibility and normalization on disconnected bounded $W^{1,2}$-extension domains.