Gaussian Poisson summation for a full-rank lattice

For a real matrix $B=(b_{ij})$ and $v\in {\mathbb{R}}^n$, each component satisfies $\mathrm{\mid }(Bv{)}_i\mathrm{\mid }\le ({\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid })\mathrm{\mid }v\mathrm{\mid }$, since $\mathrm{\mid }{v}_j\mathrm{\mid }\le \mathrm{\mid }v\mathrm{\mid }$; hence $\mathrm{\mid }Bv\mathrm{\mid }\le C_B\mathrm{\mid }v\mathrm{\mid }$ with $C_B:=\sqrt{n}{\max }_i{\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid }<\mathrm{\infty }$. Apply this to $B=A^{-1},A^T,A^{-T}$ and denote their constants by $C_{A^{-1}},C_{A^T},C_{A^{-T}}$. Define $p(y):={\sum }_{k\in {\mathbb{Z}}^n}{e}^{-\pi t\mathrm{\mid }A(y+k){\mathrm{\mid }}^2}$ for $y\in {\mathbb{R}}^n$. On any compact $K\subset {\mathbb{R}}^n$, put $M_K:={\sup }_{y\in K}\mathrm{\mid }y\mathrm{\mid }<\mathrm{\infty }$. If $\mathrm{\mid }k\mathrm{\mid }\ge 2(M_K+1)$ then $\mathrm{\mid }y+k\mathrm{\mid }\ge \mathrm{\mid }k\mathrm{\mid }-M_K\ge \mathrm{\mid }k\mathrm{\mid }\mathrm{/}2$, and therefore $\mathrm{\mid }A(y+k)\mathrm{\mid }\ge C_{A^{-1}}^{-1}\mathrm{\mid }k\mathrm{\mid }\mathrm{/}2$. The summand is bounded uniformly for $y\in K$ by $e^{-a\mathrm{\mid }k{\mathrm{\mid }}^2}$ with $a=\pi t\mathrm{/}(4C_{A^{-1}}^2)>0$. The number of integer vectors with $j\le \mathrm{\mid }k\mathrm{\mid }<j+1$ is at most $(2j+3{)}^n$, and ${\sum }_{j\ge 1}(2j+3{)}^n{e}^{-a{j}^2}$ converges by [F3], since its successive-term ratio tends to $0$. Thus the series converges absolutely and uniformly on every compact $K$. Its finite partial sums are continuous, so [F7], applied on each compact cube, makes $p$ continuous on ${\mathbb{R}}^n$. Reindexing the absolutely convergent series gives $p(y+\mathrm{\ell })=p(y)$ for every $\mathrm{\ell }\in {\mathbb{Z}}^n$.

For $m\in {\mathbb{Z}}^n$, integrate the finite partial sums of [L1] over $[0,1{]}^n$ and pass to the limit using the uniform convergence of step 1.1. For each $k$, substitution $z=y+k$ gives ${\int }_{[0,1{]}^n}{e}^{-\pi t\mathrm{\mid }A(y+k){\mathrm{\mid }}^2}{e}^{-2\pi im\cdot y}dy={\int }_{k+[0,1{]}^n}{e}^{-\pi t\mathrm{\mid }Az{\mathrm{\mid }}^2}{e}^{-2\pi im\cdot z}dz,$ because $e^{2\pi im\cdot k}=1$. The cubes for $k\in [-N,N{]}^n\cap {\mathbb{Z}}^n$ have disjoint interiors and union $[-N,N+1{]}^n$ up to their faces. Their images under $A$ form a nested compact Jordan exhaustion of ${\mathbb{R}}^n$. The integrand's modulus after the linear change $w=Az$ is $e^{-\pi t\mathrm{\mid }w{\mathrm{\mid }}^2}$, which is absolutely improperly integrable by [F2]. Thus [F8] shows that the integrals over these parallelepipeds converge to the full-space integral. Apply [F4] to the real and imaginary parts on each finite box and then take the limit. This gives $\widehat{p}(m)=c^{-1}{\int }_{{\mathbb{R}}^n}{e}^{-\pi t\mathrm{\mid }w{\mathrm{\mid }}^2}{e}^{-2\pi i(A^{-T}m)\cdot w}dw=c^{-1}{t}^{-n\mathrm{/}2}{e}^{-\pi \mathrm{\mid }{A}^{-T}m{\mathrm{\mid }}^2\mathrm{/}t}$ by [F2].

Define $s(y):=c^{-1}{t}^{-n\mathrm{/}2}{\sum }_{m\in {\mathbb{Z}}^n}{e}^{-\pi \mathrm{\mid }{A}^{-T}m{\mathrm{\mid }}^2\mathrm{/}t}{e}^{2\pi im\cdot y}$. Since $\mathrm{\mid }{A}^{-T}m\mathrm{\mid }\ge C_{A^T}^{-1}\mathrm{\mid }m\mathrm{\mid }$ by the matrix estimate in step 1.1, the same shell and ratio estimate proves absolute, uniform convergence on ${\mathbb{R}}^n$; [F7] makes $s$ continuous, and reindexing proves it is integer-periodic. Integrating its finite partial sums and using ${\int }_{[0,1{]}^n}{e}^{2\pi i(m-r)\cdot y}dy=1$ for $m=r$ and $0$ otherwise (factor the integral into one-dimensional finite integrals) shows that $\widehat{s}(m)=\widehat{p}(m)$ by step 2.1.

Put $h=p-s$. It is continuous, integer-periodic, and has every Fourier coefficient zero by step 3.1. For $N\ge 0$ define the one-dimensional Fejér kernel $F_N(u):=\frac{1}{N+1}{\mid {\sum }_{j=0}^N{e}^{2\pi iju}\mid }^2={\sum }_{\mathrm{\mid }r\mathrm{\mid }\le N}(1-\frac{\mathrm{\mid }r\mathrm{\mid }}{N+1})e^{2\pi iru},$ and the product kernel $K_N(z):={\prod }_{j=1}^n{F}_N(z_j)$. Expanding the square shows $F_N\ge 0$ and gives the displayed finite Fourier sum; its integral on $[0,1]$ is its constant coefficient $1$, so $K_N\ge 0$ and its integral on $[0,1{]}^n$ is $1$. Write $K_N(z)={\sum }_{m\in [-N,N{]}^n}{w}_m{e}^{2\pi im\cdot z}$. For each term, the substitution $u=y-z$ gives ${\int }_{[0,1{]}^n}h(y-z)e^{2\pi im\cdot z}dz=e^{2\pi im\cdot y}{\int }_{y-[0,1{]}^n}h(u)e^{-2\pi im\cdot u}du=e^{2\pi im\cdot y}\widehat{h}(m)\mathrm{.}$ The last equality holds because $u\mapsto h(u)e^{-2\pi im\cdot u}$ is continuous and $1$-periodic in each coordinate; its integral is unchanged when a unit interval is translated, by splitting the translated interval at an integer and using periodicity, and applying this one-coordinate identity successively to the rectangular integral. Hence ${\sigma }_{N}h(y):={\int }_{[0,1{]}^n}h(y-z)K_N(z)dz={\sum }_m{w}_m{e}^{2\pi im\cdot y}\widehat{h}(m)=0$. To see that ${\sigma }_{N}h\to h$ uniformly, first use periodicity to take $y\in [0,1{]}^n$. Apply [F6] to $h$ on the compact cube $[-1,2{]}^n$ to choose $0<\delta <1\mathrm{/}2$ such that $\mathrm{\mid }h(y-w)-h(y)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }w{\mathrm{\mid }}_{\mathrm{\infty }}<\delta$ and both points lie in that cube. If every coordinate of $z$ has distance less than $\delta$ from $\mathbb{Z}$, choose $k\in {\mathbb{Z}}^n$ with $w=z-k$ and $\mathrm{\mid }w{\mathrm{\mid }}_{\mathrm{\infty }}<\delta$; periodicity gives $h(y-z)=h(y-w)$. On the complement, the geometric-sum formula ${\sum }_{j=0}^N{e}^{2\pi iju}=(1-e^{2\pi i(N+1)u})\mathrm{/}(1-e^{2\pi iu})$ and $\mathrm{\mid }1-e^{2\pi iu}\mathrm{\mid }=2\mathrm{\mid }\sin \pi u\mathrm{\mid }$ give $F_N(u)\le ((N+1){\sin }^2(\pi \delta ){)}^{-1}$. The complement is a union of $n$ coordinate-tail sets; integrating the product kernel, using nonnegativity and the unit integral of each other coordinate factor, bounds its mass by $n\mathrm{/}((N+1){\sin }^2(\pi \delta ))$. The uniform majorants in steps 1.1 and 3.1 bound $\mathrm{\mid }h\mathrm{\mid }$ by some finite $M$, so ${\sup }_y\mathrm{\mid }{\sigma }_{N}h(y)-h(y)\mathrm{\mid }\le \epsilon +2\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{n}{(N+1){\sin }^2(\pi \delta )},$ which tends to at most $\epsilon$. Since $\epsilon$ is arbitrary and every ${\sigma }_{N}h=0$, $h=0$ and $p=s$.

For arbitrary $x\in {\mathbb{R}}^n$ set $y=A^{-1}x$. Then $p(y)={\sum }_{k\in {\mathbb{Z}}^n}{e}^{-\pi t\mathrm{\mid }x+Ak{\mathrm{\mid }}^2}$. In the series for $s(y)$ put $\mu =A^{-T}m$; then $m\cdot y=(A^{-T}m)\cdot x=\mu \cdot x$. The equality $p=s$ from step 4.1 is therefore exactly the identity in the Statement. The uniform absolute convergence proved in steps 1.1 and 3.1 establishes the convergence of both coordinate-box sums.