Lemma: Transfer singularity and polynomial bound

Put

There is $C_f<\infty$ such that

For a selected stress pair:

1. equal signs make its complete transfer numerator $O(r^2)$, so its kernel is locally bounded at $r=0$;

2. opposite signs generally leave a nonzero $r^{-2}$ coefficient, which is locally integrable in three transverse dimensions.

No additional frequency pole occurs in the finite-time kernels.

Proof. Use transverse coordinates

This linear transformation has unit absolute Jacobian. For equal signs,

by expanding ${\omega }_{-\mathbf{p}+\sigma \mathbf{q}}={\omega }_{\mathbf{p}}-\sigma \mathbf{p}\cdot \mathbf{q}/{\omega }_{\mathbf{p}}+O(r^2)$ in (eq:numerators). Both initial terms in (eq:Z) are then bounded; the source term is smooth. The divergence terms are bounded by (eq:dcancel).

For opposite signs, at $\mathbf{q}=0$,

The temporal initial amplitude gives a leading coefficient

where

The spatial initial term contributes at most an $r^{-1}$ term. The source term is smooth and cannot cancel the displayed constant numerator universally. Small-time nonnegative smearings, with generic momenta, make $\mathcal{I}(0)\ne 0$. Away from the other pair transfers only this cyclic summand has the pole, so symmetrization does not universally cancel it.

For the global bound, on the finite time interval

The stress numerator is $O(R^2)$, the spatial initial numerator is $O(R^3)$, and the contraction with $k_3{k}_3$ costs at most $C{R}^2$. The divergence terms cost at most $C{R}^3$; the massive mode factors are bounded. Also $|F_f(t,\mathbf{w})|\le \Vert f(t,\cdot ){\Vert }_1$. These facts give $C{R}^5(1+r^{-2})$ for each ordered pair and (eq:cubicmajorant) after summing.

All oscillatory factors are retained as finite time integrals. For example $(e^{i\nu L}-1)/(i\nu )$ means ${{\int }_0}^L{e}^{i\nu s}\mathrm{d}s$, continuous at $\nu =0$. No split resonance denominator is introduced.