The L2 pivot injects into the local negative Sobolev space

The test functions form a linear subspace of $H^2(G)$: $0$ is a test function, and $a\phi +b\psi$ is smooth with support contained in the compact union $\mathrm{supp}\phi \cup \mathrm{supp}\psi \subset G$ for scalars $a,b$ and test functions $\phi ,\psi$. Let $u,v\in V$, $a,b\in \mathbb{K}$ and $\epsilon >0$. By the closure criterion in [F1], choose test functions $\phi ,\psi$ such that $\mathrm{\parallel }u-\phi {\mathrm{\parallel }}_{H^2},\mathrm{\parallel }v-\psi {\mathrm{\parallel }}_{H^2}<\epsilon \mathrm{/}(1+\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid })$. The norm axioms give $\mathrm{\parallel }au+bv-(a\phi +b\psi ){\mathrm{\parallel }}_{H^2}\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }u-\phi {\mathrm{\parallel }}_{H^2}+\mathrm{\mid }b\mathrm{\mid }\mathrm{\parallel }v-\psi {\mathrm{\parallel }}_{H^2}<\epsilon$. Thus $au+bv\in V$. Also $0\in V$, so $V$ is a linear subspace; it is closed because it is the metric closure of the test functions.

By [F4], $(Jf)(\phi )=⟨f,\phi ⟩$ is independent of the representatives of $f$ and $\phi$, is additive in $\phi$, and has the asserted linear or conjugate-linear scalar dependence. By [F3] and [F1], $\mathrm{\mid }(Jf)(\phi )\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }\phi {\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }\phi {\mathrm{\parallel }}_V$. Thus it belongs to $Y$. Its dependence on $f$ is linear by [F4], and taking the supremum over the unit ball of $V$ proves $\mathrm{\parallel }Jf{\mathrm{\parallel }}_Y\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$.

By [F1] and step 1.1, $V$ is a closed linear subspace of the Banach space $H^2(G)$, so the closed-subspace theorem proves $V$ Banach. In the real case [F2] proves $Y$ Banach. In the complex case the map $F\mapsto \bar{F}$, with $(\bar{F})(\phi )=\overline{F(\phi )}$, is a distance-preserving bijection from the conjugate dual to the ordinary linear dual. A Cauchy sequence in $Y$ therefore has a limit there, since its conjugates converge in the ordinary dual and the inverse conjugation preserves distances.

If $Jf=0$, choose ${\phi }_m\in C_c^{\mathrm{\infty }}(G)\subset V$ with $\mathrm{\parallel }f-{\phi }_m{\mathrm{\parallel }}_2<1\mathrm{/}m$ for $m\ge 1$, using [F3] and Countable Choice from the assumed Axiom of Choice. Each $⟨f,{\phi }_m⟩=(Jf)({\phi }_m)$ vanishes. By additivity and the diagonal identity in [F4], $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=⟨f,f⟩=⟨f,f-{\phi }_m⟩$. The bound in [F3] gives $\mathrm{\mid }⟨f,f-{\phi }_m⟩\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }f-{\phi }_m{\mathrm{\parallel }}_2\to 0$. Hence $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=0$, so $f=0$ as an $L^2$ class and $J$ is injective.