Bounded H1 functionals have compatible local L2 representatives

For a cube $Q$, the set $L_0^2(Q)$ is the kernel of the continuous linear functional $f\mapsto {\int }_{Q}f$ on the closed subspace $\{f\in L^2:f=0\text{ a.e. off }Q\}$ (closedness follows from $\mathrm{\parallel }f{\mathbf{1}}_{Q^c}{\mathrm{\parallel }}_2\le \mathrm{\parallel }f-f_m{\mathrm{\parallel }}_2$ for a supported approximating sequence $f_m$, and continuity of the integral from [F4]), hence a closed subspace of the Hilbert space $L^2({\mathbb{R}}^n)$; and for $f\in L_0^2(Q)$ the boundedness of $\mathrm{\Lambda }$ and [F1] give $\mathrm{\mid }\mathrm{\Lambda }(f)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}\le C_{n,\tilde{N},\phi }\mathrm{\mid }Q{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}$, so $\mathrm{\Lambda }{\mathrm{\mid }}_{L_0^2(Q)}$ is bounded for the $L^2$ norm.

By [F2] applied to the closed subspace $L_0^2(Q)$ and the bounded functional $\mathrm{\Lambda }{\mathrm{\mid }}_{L_0^2(Q)}$, there is a unique $y_Q\in L_0^2(Q)$ with $\mathrm{\Lambda }(f)=⟨f,y_Q⟩$ for $f\in L_0^2(Q)$; setting $u_Q:=\overline{y_Q}$, which still lies in $L_0^2(Q)$ because conjugation preserves supports and means, gives $\mathrm{\Lambda }(f)={\int }_Q{u}_{Q}f$ for every $f\in L_0^2(Q)$, and $u_Q$ is unique with this property.

Nested compatibility. If $\mathrm{\mid }Q\mathrm{\mid }=0$, then $L_0^2(Q)=\{0\}$ as an almost-everywhere quotient, and constancy almost everywhere on $Q$ is vacuous. Assume now $\mathrm{\mid }Q\mathrm{\mid }>0$. If $Q\subseteq R$ then $L_0^2(Q)\subseteq L_0^2(R)$, and for $f\in L_0^2(Q)$ step 2.1 gives ${\int }_Q{u}_{Q}f=\mathrm{\Lambda }(f)={\int }_R{u}_{R}f={\int }_Q{u}_{R}f$. Put $h:=u_Q-u_R$ and apply this identity with $f:=\overline{h-{\mathrm{mean}}_Q(h)}{\mathbf{1}}_Q$, which belongs to $L_0^2(Q)$. Then ${\int }_Q\mathrm{\mid }h-{\mathrm{mean}}_Q(h){\mathrm{\mid }}^2=0$, so $u_Q-u_R$ equals the constant ${\mathrm{mean}}_Q(u_Q-u_R)$ almost everywhere on $Q$.

Gluing. Let $Q_k=[-k,k{]}^n$ for $k\ge 1$ and use Countable Choice to select measurable representatives of their $u_{Q_k}$. By step 3.1 the difference $(u_{Q_{k+1}}-u_{Q_k})$ is almost everywhere constant $a_k$ on $Q_k$; define $c_1:=0$ and $c_{k+1}:=c_k-a_k$, so that $u_{Q_{k+1}}+c_{k+1}=u_{Q_k}+c_k$ almost everywhere on $Q_k$. Removing the countable union of the exceptional null sets, which is null by [F3], define $u(x):=u_{Q_k}(x)+c_k$ for $x\in Q_k$ outside that null set, and set $u=0$ on the null set; this is well defined, locally integrable, and for every cube $Q$, choosing $k$ with $Q\subseteq Q_k$, the function $u-u_Q=[u-(u_{Q_k}+c_k)]+[(u_{Q_k}+c_k)-u_Q]$ is almost everywhere constant on $Q$ by the construction and step 3.1. If $u^{\mathrm{\prime }}$ is another such function, then on each $Q_k$ the difference $u-u^{\mathrm{\prime }}$ is constant almost everywhere, and the constants agree on the positive-measure overlap $Q_k\cap Q_{k+1}=Q_k$, so $u-u^{\mathrm{\prime }}$ is almost everywhere equal to a single constant on ${\bigcup }_k{Q}_k={\mathbb{R}}^n$: uniqueness up to additive constants.

Steps 2.1, 3.1 and 4.1 prove the existence and uniqueness of each $u_Q$, the nested constancy, and the existence of the global representative $u$ with constants $c_Q$, which is the statement. Countable Choice is used for the countably many representations and the countable union in step 4.1.