Fixed-regularization periodic PIC force consistency on a smooth density

The assigned source has mean zero: by the partition identity in [F1], $h^d{\sum }_j(P_h^{\epsilon }\mu {)}_j=\int {\sum }_j{w}_j(x)d\mu (x)=Q$, so subtracting the background gives zero mean. By [F4], the continuum potential and field are smooth; [F2] gives the unique grid potential. The assignment and centered-stencil Taylor bounds give the residual $r_h=s_h-A_h({\varphi }_{\epsilon }{\mathrm{\mid }}_h),\mathrm{\parallel }{r}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\mathrm{s}\mathrm{r}\mathrm{c}}{h}^2\mathrm{.}$ Here [F1] bounds the measure-assignment error by $C_{\mathrm{d}\mathrm{e}\mathrm{p}}{h}^2$, and for each coordinate the centered second difference has remainder at most $h^2\mathrm{\parallel }{\mathrm{\partial }}_a^4{\varphi }_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$, whose sum is $C_{\mathrm{F}\mathrm{D}}{h}^2$.

The weighted grid norm satisfies $\mathrm{\parallel }{r}_h{\mathrm{\parallel }}_h\le L^{d\mathrm{/}2}\mathrm{\parallel }{r}_h{\mathrm{\parallel }}_{\mathrm{\infty }}$ because the grid has $N^d$ points and $h^d{N}^d=L^d$. In [F3], take $s=s_h$, $\varphi ={\varphi }_h$, and $\tilde{\varphi }={\varphi }_{\epsilon }{\mathrm{\mid }}_h$; its residual is exactly $r_h$ from step 1.1, so $\mathrm{\parallel }{D}^0({\varphi }_h-{\varphi }_{\epsilon }{\mathrm{\mid }}_h){\mathrm{\parallel }}_h\le {\gamma }_{N,L}{L}^{d\mathrm{/}2}{C}_{\mathrm{s}\mathrm{r}\mathrm{c}}{h}^2$. Taylor's theorem for the centered first difference gives $\mathrm{\parallel }{D}^0({\varphi }_{\epsilon }{\mathrm{\mid }}_h)-(\mathrm{\nabla }{\varphi }_{\epsilon }){\mathrm{\mid }}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_3{h}^2$, hence its grid $L^2$ norm is at most $L^{d\mathrm{/}2}{C}_3{h}^2$.

For any nonnegative normalized weights, ${\sum }_j{w}_j(x{)}^2\le 1$, so Cauchy–Schwarz gives $\mathrm{\mid }{G}_{x}V\mathrm{\mid }\le h^{-d\mathrm{/}2}\mathrm{\parallel }V{\mathrm{\parallel }}_h$. Apply this to the two grid-field errors in step 2.1. Their gathered contribution is at most $L^{d\mathrm{/}2}{h}^{2-d\mathrm{/}2}({\gamma }_{N,L}{C}_{\mathrm{s}\mathrm{r}\mathrm{c}}+C_3)$.

The remaining difference is gathering the exact nodal samples $E_{\epsilon }(x_j)$ instead of the continuous field. Apply the Euclidean-vector bound of [F1] with $m=d$ and $g=E_{\epsilon }$. Its translation-uniform constant is exactly $C_{\eta ,E}$ in the Statement, so this difference is at most $C_{\mathrm{i}\mathrm{n}\mathrm{t}}{h}^2$. The unit-vector projection proof in that supplier introduces no $\sqrt{d}$ factor. Add this to step 3.1 to get the displayed estimate. Since ${\gamma }_{N,L}\le L\mathrm{/}4$ and $2-d\mathrm{/}2>0$ for $d\le 3$, both terms tend to zero.