Poisson summation under two-sided polynomial decay

Fix a compact set $K$, let $R$ bound $\mathrm{\mid }x\mathrm{\mid }$ on $K$, and choose $j_0$ so $2^j\ge 2R$ for $j\ge j_0$. For $2^j\le \mathrm{\mid }\lambda \mathrm{\mid }<2^{j+1}$ we have $\mathrm{\mid }x+\lambda \mathrm{\mid }\ge 2^{j-1}$, so $\mathrm{\mid }f(x+\lambda )\mathrm{\mid }\le C^{\mathrm{\prime }}{2}^{-j(n+\epsilon )}$ uniformly on $K$. The annulus contains at most the number of lattice points in the ball of radius $2^{j+1}$, hence at most $C^{\mathrm{\prime }\mathrm{\prime }}{2}^{jn}$ by [F1]; its total contribution is therefore $O(2^{-j\epsilon })$. These contributions are summable by [F3]. The finitely many lattice points with $\mathrm{\mid }\lambda \mathrm{\mid }<2^{j_0}$ contribute a finite sum of continuous functions, uniformly on $K$. Thus $P(x):={\sum }_{\lambda \in \mathrm{\Lambda }}f(x+\lambda )$ converges absolutely and uniformly on $K$; its real and imaginary parts are uniform limits of continuous real-valued functions, so [F4] makes both limits continuous and [F9] makes $P$ continuous on $K$. Since $K$ was arbitrary, $P$ is continuous on ${\mathbb{R}}^n$, and reindexing the absolutely convergent series gives $P(x+\mu )=P(x)$ for $\mu \in \mathrm{\Lambda }$.

By the tiling [F2], Tonelli and the integrability of $f$, ${\sum }_{\lambda \in \mathrm{\Lambda }}{\int }_F\mathrm{\mid }f(x+\lambda )\mathrm{\mid }dx={\int }_{{\mathbb{R}}^n}\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$; hence [F6] justifies termwise integration and, substituting $y=x+\lambda$ with $e^{-2\pi i{\lambda }^{\ast }\cdot (y-\lambda )}=e^{-2\pi i{\lambda }^{\ast }\cdot y}$ [F5], ${\int }_{F}P(x)e^{-2\pi i{\lambda }^{\ast }\cdot x}dx={\sum }_{\lambda }{\int }_{F+\lambda }f(y)e^{-2\pi i{\lambda }^{\ast }\cdot y}dy=\widehat{f}({\lambda }^{\ast })$ for every ${\lambda }^{\ast }\in {\mathrm{\Lambda }}^{\ast }$.

Define $S(x):=c^{-1}{\sum }_{{\lambda }^{\ast }\in {\mathrm{\Lambda }}^{\ast }}\widehat{f}({\lambda }^{\ast })e^{2\pi i{\lambda }^{\ast }\cdot x}$. Grouping the dual lattice into dyadic annuli and applying [F1], each annulus contributes $O(2^{-j\epsilon })$ by the decay bound on $\widehat{f}$; [F3] makes the series absolutely convergent. Since each exponential has modulus one, this convergence is uniform on ${\mathbb{R}}^n$; each summand is continuous by [F8], so its real and imaginary partial sums converge uniformly to continuous functions, and [F4] and [F9] make $S$ continuous. It is $\mathrm{\Lambda }$-periodic because ${\lambda }^{\ast }\cdot \lambda \in \mathbb{Z}$ for ${\lambda }^{\ast }\in {\mathrm{\Lambda }}^{\ast }$, $\lambda \in \mathrm{\Lambda }$. Integrating term by term with [F5] and [F6] gives ${\int }_{F}S(x)e^{-2\pi i{\eta }^{\ast }\cdot x}dx=c^{-1}{\sum }_{{\lambda }^{\ast }}\widehat{f}({\lambda }^{\ast }){\int }_F{e}^{2\pi i({\lambda }^{\ast }-{\eta }^{\ast })\cdot x}dx=\widehat{f}({\eta }^{\ast })$ for every ${\eta }^{\ast }\in {\mathrm{\Lambda }}^{\ast }$.

By steps 2.1 and 2.2 the continuous $\mathrm{\Lambda }$-periodic function $P-S$ has every lattice Fourier coefficient zero, so $P=S$ by [F7]; this is the displayed identity, and its left side converges absolutely by step 1.1. Evaluating at $x=0$ gives ${\sum }_{\lambda }f(\lambda )=c^{-1}{\sum }_{{\lambda }^{\ast }}\widehat{f}({\lambda }^{\ast })$. The two-sided $(n+\epsilon )$-decay hypotheses are used only through the majorants of steps 1.1 and 2.2; they cannot be dropped to bare integrability with point values, as recorded on the companion page. Countable Choice is inherited from the integration and convergence suppliers above.