Theorem: Evaluated complete-family decision

For the fixed specialization above,

The bound holds uniformly on arbitrary tests supported in each fixed $L⋐U$, for all sufficiently small scales.

The apparent ${\delta }^{-2}$ coefficient is zero. The complete inverse-scale coefficient is (eq:appA), and the finite part is (eq:appT). Both are supported in $S_{\kappa }$, and $\mathcal{A}\ne 0$.

Consequently one fixed compact joint test has an unbounded raw pairing, whereas

strongly, with the displayed order-six $O(\delta )$ rate.

Proof. Apply Lemma lem:angular to the $j$-term of (eq:appintegral), with $M=2-j$. The Taylor error is

Multiplication by ${\delta }^{j-2}$ leaves $O(\delta )$. Here $F_0$ contains no ambient test derivative, while $F_1,F_2$ contain at most one. Their coefficients and coordinate derivatives are bounded on the relevant compact range. Hence ambient test order six controls every remainder. The fixed compact integration ranges preserve uniformity.

At order ${\delta }^{-2}$, only $j=0,n=0$ contributes. Its moment is $M_{2,0,0}=0$. At order ${\delta }^{-1}$, the pairs $(j,n)=(0,1),(1,0)$ give (eq:appA). At finite order, the pairs $(0,2),(1,1),(2,0)$ give (eq:appT).

The formulas are continuous finite-jet functionals along $S_{\kappa }$, so they define distributions supported there.

For nonvanishing, choose nonnegative nonzero $\Psi (\tau )$ compactly supported in a small tangential subchart. Choose a compact normal cutoff equal to one near $\nu =0$, and let ${\Phi }_{\ast ,0}$ be their product, with all other components zero. The support can be chosen inside $U$. Along the attachment,

because ${\partial }_{b^0}=({\partial }_B+{\partial }_u)/2$. Tangential integration kills ${\partial }_B\Psi$. Therefore

The coefficient in parentheses is positive on the patch.

Thus, with $c=\mathcal{A}({\Phi }_{\ast })\ne 0$,

The complete-expansion theorem gives the claimed raw failure and strong subtracted convergence.