Proposition: Coverage and an explicit finite-test limitation

Let $\mathcal{V}$ be an independently specified finite-dimensional space containing all residuals admissible for a particular claim. If a finite evaluation map

is injective, then $\mathcal{Q}(R)=0$ implies $R=0$ within that declared class.

Without a coverage hypothesis, finite tests do not in general establish a distributional identity. In particular, any finite family of compact linear evaluations of smooth symmetric tensors can miss a nonzero compactly supported conserved perturbation in a regular developed region.

Proof. The first assertion is injectivity.

For the second, work in a relatively compact regular Minkowski chart and, for $h\in {C_c}^{\infty }$ there, put

This tensor is symmetric and conserved. Every one of the finitely many evaluations is a linear functional of $h$. Their common kernel in the infinite-dimensional space of such $h$ contains a nonzero element. Its trace is

A compactly supported smooth solution of ${\square }_{\eta }h=0$ has zero Cauchy data before its time support and is zero. Hence the chosen nonzero $h$ has nonzero trace, and $S[h]\ne 0$. Extension by zero preserves smoothness because its support lies strictly inside the regular chart.