Summability and convolution control for lattice weights

For $k\ge 1$, the $d_{\mathrm{\infty }}$-shell of radius $k$ has $(2k+1{)}^d-(2k-1{)}^d\le c_d(k+1{)}^{d-1}$ sites, by factoring the difference of powers; the radius-zero shell has one site. Therefore ${\sum }_z{F}_{\rho }(d_{\mathrm{\infty }}(x,z))\le c_d{\sum }_{k\ge 0}(k+1{)}^{d-1-\rho }<\mathrm{\infty }$, and ${\sum }_z{F}_a(d_{\mathrm{\infty }}(x,z))\le {\sum }_{z}F(d_{\mathrm{\infty }}(x,z))<\mathrm{\infty }$. The bounds are independent of $x$ by translation invariance.

Write $r=d_{\mathrm{\infty }}(x,y)$. For $G=F_{\rho }$, the triangle inequality places every $z$ in at least one of the sets $d_{\mathrm{\infty }}(x,z)\ge r\mathrm{/}2$ or $d_{\mathrm{\infty }}(z,y)\ge r\mathrm{/}2$. On the first set, $G(d_{\mathrm{\infty }}(x,z))\le 2^{\rho }G(r)$; summing the other factor over all $z$ is bounded by the finite shell sum in step 1.1. The second set gives the same bound with $x,y$ exchanged. Hence $C_{F_{\rho }}\le 2^{\rho +1}{\sum }_z{F}_{\rho }(d_{\mathrm{\infty }}(0,z))<\mathrm{\infty }$. For $F_a$, the triangle inequality gives $e^{-a(d(x,z)+d(z,y))}\le e^{-ar}$; the same splitting argument applied to $F=F_{d+1}$, for which $d+1>d$, yields $C_F<\mathrm{\infty }$, and then ${\sum }_z{F}_a(d(x,z))F_a(d(z,y))\le C_F{F}_a(r)$. Positivity of each weight and the $z=x$ summand show $C_G>0$.