Lemma: Low-frequency and direct sectors

The distribution $S$ exists, is independent of $\rho$, and satisfies

Proof. The profiles are bounded and globally Lipschitz, with value one at zero. Telescoping the product of multipliers gives

For $|q|\le 1$, (eq:FTbound) and the bounded-$r$ part of Lemma lem:convolution give an integrable bound. The extra powers in (eq:multdifference) are harmless: $P\le |\alpha |+1$ and $P^{\prime }\le |\beta |+1$. Thus this sector has a cutoff-free limit with $O(\epsilon )$ error.

For $r=|q|\ge 1$, suppose the left branch is direct. Its $t$-frequency is $-(P+{\omega }_m(q))$. Integrating eight times by parts in $t$ gains

The amplitude is at most linear in $t$; its differentiated coefficients are still bounded by $C_K(P^{-1}+P^{-2})$. The other amplitude is bounded by $C_K(P^{\prime -1}+P^{\prime -2})$. If only the right branch is direct, use $t^{\prime }$ instead.

The shifted bounds show that the two momentum integrations contribute at most $C{r}^{-2}$. Since $r^2{\omega }_m(r)\le C_m{r}^3$, the unregulated radial majorant is $C{r}^{-7}$. For the regulated difference,

and (eq:multdifference) costs at most $C\epsilon r$ after the weighted integrations. The resulting radial majorant is $C\epsilon r^{-6}$. Both are integrable.

All integrations by parts are performed against the full compact joint test. No restriction to an observation diagonal is used.