Lemma: Finite-part formula for the component kernels

The limit

exists and is independent of the profile. The sum is empty when $p=2$. It defines a distribution of transverse scaling degree at most $p$, equal to ${f_A}^{\circ }$ off $\mathcal{W}$.

For the scalar, stress, and vector kernels below, parity improves the fixed-support convergence remainder to $O({\epsilon }^2)$.

Proof. Expand the test transversely through degree $p-2$. Radial integration of the $k$-th term gives the power ${\epsilon }^{k+2-p}$ when $k<p-2$, and a logarithm when $k=p-2$. These are precisely the subtractions in (eq:rawfinitepart). The Taylor remainder is integrable.

Comparing two profiles on their transition annuli gives the same moment contributions, so the normalized limit is profile independent. Taylor remainder estimates prove continuity and the stated scaling-degree bound.

For $r^{-2}$, angular odd terms vanish. For the stress kernels the angular coefficient is even, so the cubic Taylor term vanishes. For $x_i/r^4$, the angular coefficient is odd, so the quadratic Taylor term vanishes. In these cases the first unremoved angular term yields an $O({\epsilon }^2)$ remainder, bounded respectively by test seminorms of orders $2,4,3$.