Proposition: Coverage in a declared defect space

Let $\mathcal{V}$ be an independently specified finite-dimensional space containing every admissible residual distribution for the claim being tested. If

is injective, then $E(W)=0$ implies $W=0$ in $\mathcal{V}$.

Without such a coverage condition, vanishing of the finite evaluations does not generally imply a distributional identity.

Proof. The first assertion is injectivity. For the second, any nonzero element of $\mathcal{V}\cap \ker E$, when it exists, is invisible to every declared evaluation.