Poisson summation for Schwartz functions

For each $\beta$, expansion of $W(y)={\prod }_j(1+y_j^2)$ gives $\mathrm{\mid }{\mathrm{\partial }}^{\beta }f(y)\mathrm{\mid }\le A_{\beta }\mathrm{/}W(y)$ with $A_{\beta }={\sum }_{ϵ\in \{0,1{\}}^n}{p}_{2ϵ,\beta }(f)$, as in [F2]. On a fixed bounded box $\mathrm{\mid }{x}_j\mathrm{\mid }\le L$, the elementary inequality $1+k_j^2\le (2+2L^2)(1+(x_j+k_j{)}^2)$ shows $\mathrm{\mid }{\mathrm{\partial }}^{\beta }f(x+k)\mathrm{\mid }\le A_{\beta }(2+2L^2{)}^n{\prod }_j(1+k_j^2{)}^{-1}$. The one-dimensional series of reciprocal weights converges: on $2^r\le \mathrm{\mid }k\mathrm{\mid }<2^{r+1}$ its sum is at most $2^{1-r}$, with the term $k=0$ separate. The finite product series therefore converges. Uniform tail bounds prove absolute uniform convergence of all derivative series on that box. Applying [F5] componentwise on each coordinate segment to finite partial sums, and iterating for ordered derivatives, proves that $P(x)={\sum }_{k}f(x+k)$ is smooth with the asserted derivatives. Absolute convergence permits reindexing, so $P$ is periodic.

On $Q=[0,1{)}^n$, the tiling $\{Q+k:k\in {\mathbb{Z}}^n\}$ is disjoint and exhausts ${\mathbb{R}}^n$. Translation substitution [F7] and [F4] give ${\sum }_k{\int }_Q\mathrm{\mid }f(x+k)\mathrm{\mid }dx=\int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$. Thus exchanging the coefficient integral and sum is justified. For $\mathrm{\ell }\in {\mathbb{Z}}^n$, substitute $y=x+k$ in each term; $e^{2\pi i\mathrm{\ell }\cdot k}=1$, so ${\int }_{Q}P(x)e^{-2\pi i\mathrm{\ell }\cdot x}dx={\int }_{{\mathbb{R}}^n}f(y)e^{-2\pi i\mathrm{\ell }\cdot y}dy=\widehat{f}(\mathrm{\ell })$. The closed cube gives the same integral because its added coordinate faces are null.

By [F1] and the weight estimate of step 1.1 at $x=0$, ${\sum }_k\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }<\mathrm{\infty }$. Therefore $H(x)={\sum }_k\widehat{f}(k)e^{2\pi ik\cdot x}$ converges absolutely uniformly for all $x$, is continuous and periodic. Its coefficient at $\mathrm{\ell }$ is $\widehat{f}(\mathrm{\ell })$: interchange sum and integral by the summable constant majorant, using [F4]. Each exponential product integral factors by [F4], and each factor is ${\int }_0^1{e}^{2\pi irt}dt=0$ for nonzero integer $r$, by its antiderivative and [F6], or $1$ for $r=0$. Hence only $k=\mathrm{\ell }$ survives.

The continuous periodic function $P-H$ has every coefficient zero by steps 2.1 and 2.2. [F3] makes it zero everywhere. Evaluation at zero gives the unshifted formula, with absolute convergence already proved. Countable choice is precisely the inherited Euclidean integration and Schwartz Fourier hypothesis; the lattice ordering, majorants and partial sums are explicit.