Lemma: Closed-range test

For $x=(A,B,u,v)\in {\mathcal{X}}_{\mathrm{C}}$,

When the equality on the right holds, all reconstructed fields are compact. The gauge range is closed in ${\mathcal{X}}_{\mathrm{C}}$.

Proof. If $x=\mathfrak{g}(\alpha ,\beta )$ for compact gauge data, then

Newton inversion recovers $\alpha$. The remaining reconstruction equations recover $\beta$.

Conversely, assume the displayed membership equality. Outside a ball containing the metric-data support,

Differentiating and combining index permutations gives

The exterior is connected, so ${\alpha }_i$ is affine there. Decay makes it zero.

Also ${\beta }_0=0$, while $A_{0i}=0$ gives ${\beta }_i=-{\partial }_i{\alpha }_0$. The exterior equation $B_{ij}=0$ then yields ${\partial }_i{\partial }_j{\alpha }_0=0$. The decaying affine ${\alpha }_0$ is zero there, as are the remaining ${\beta }_i$. All reconstructed fields are compact. The assumed equality also supplies the scalar components, so $x$ lies in the gauge range.

For closedness, Newton convolution is continuous from each compact smooth support stage to $C^{\infty }$ with its compact-open topology. Move derivatives onto the compact source and use local integrability of the Newton kernel to obtain the seminorm estimates. Hence

is continuous on each ${{\mathcal{X}}_{\mathrm{C}}}_n$, and therefore on ${\mathcal{X}}_{\mathrm{C}}$, with values in a finite product of Hausdorff smooth spaces. The gauge range is its kernel and is closed.