Lemma: Transfer-momentum cancellation

At fixed finite $\Lambda$, $g_{\Lambda }$ is a symmetric Hilbert–Schmidt pair tensor, and $h_{\Lambda }$ is a bounded self-adjoint one-particle operator. On the cutoff support,

No separate Newton infrared cutoff is needed.

Proof. For the pair sector,

The first numerator and its first $q$-derivatives vanish at $q=-p$; the second numerator vanishes there. Taylor's theorem on the bounded cutoff support gives

The same holds for $--$. Thus $A_{00}=O(1)$, $A_{0i}=O(r^{-1})$, and $B_{00}=O(1)$. Use

on the compact time support of $J$. This proves the pair bound. Its square is locally integrable in the three-dimensional variable $p+q$, since ${\int }_{|k|<1}|k{|}^{-2}\mathrm{d}k<\infty$.

For mixed signs use $A=O(r^{-2})$, $B=O(r^{-1})$, without assuming a pair cancellation. The same estimates prove the mixed bound. Its Schur integrals are finite uniformly on the bounded support, because $|p-q{|}^{-2}$ is locally integrable in three dimensions. Therefore it defines a bounded operator. Reality of $J$ and the response gives $h_{\Lambda }(p,q)=\overline{h_{\Lambda }(q,p)}$; polarized symmetry gives symmetry of $g_{\Lambda }$.