Theorem: Whole-support radial identity

At sufficiently small positive scales,

The right side is an averaged distributional pairing. The radial amplitudes are smooth up to $t=0$, have common compact radial support, and satisfy uniform finite-order estimates.

Proof. The mode conventions in Appendix app:inputs give, on the separated relative region,

The series for $h_m$ obeys $4z{h}_m^{\prime \prime }+8h_m^{\prime }+m^2{h}_m=0$ and $4h_m(0)=-m^2/(2\pi )$, matching the causal normalization. At $m=0$, it vanishes identically.

For the cone part,

Its delta sets

Completion of the square gives

The negative multiplier changes $-i0$ to $+i0$. Before angular integration the coefficient is $-1/(32{\pi }^3{e}_{\delta })$. The substitution $\xi -\delta Z_{\delta }=\sqrt{t}{\omega }_{\phi }$, with ${\mathrm{d}}^2\xi =\frac{1}{2}\mathrm{d}t\mathrm{d}\phi$, gives the cone term of (eq:radial).

For the tail, put $z=q_{2,\delta }\ge 0$. Then

and

The tail integral is

Change to $w=t+c_{\delta }z$. The inner Jacobian is $1/c_{\delta }$, and the ratio of tail and cone prefactors is $2\pi$. This yields (eq:radial-amplitude).

Circular averaging eliminates odd Cartesian Taylor terms and is smooth in the squared radius. A radial derivative costs at most two Cartesian amplitude derivatives. The corresponding integral remainders are uniform on compact parameter sets.

Compact amplitude support bounds $\xi ,v$. The displayed substitutions bound both nonnegative radial variables. Thus the collapsed amplitude has common compact support.

For the distributional justification, replace $i0$ by $i\epsilon$, perform the ordinary changes, pair in $H$, and let $\epsilon \downarrow 0$. The map $H\mapsto L_{\delta }$ is a submersion in $B_2-B_1$. The averaged boundary values therefore exist without assigning an individual-displacement crossed product.