Muhammad Haris Rao
Recall the definition of the Schwartz space:
Definition: We say that a smooth function $f : \mathbb{R} \longrightarrow \mathbb{C}$ is rapidly decreasing if for every $n, m \in \mathbb{Z}_{\ge 0}$, \begin{align*} \sup_{x \in \mathbb{R}} \left| x^n f^{(m)} (x) \right| < \infty \end{align*} The Schwartz space $\mathcal{S}(\mathbb{R})$ is \begin{align*} \mathcal{S} \left( \mathbb{R} \right) &= \left\{ f \in \mathcal{C}^\infty \left( \mathbb{R} \right) \mid \text{$f$ is rapidly decreasing} \right\} \end{align*}
For $f \in \mathcal{S} \left( \mathbb{R} \right)$, its periodisation $F ; \mathbb{R} \longrightarrow \mathbb{C}$ is defined as \begin{align*} F \left( x \right) &= \sum_{n \in \mathbb{Z}} f(x + n) \end{align*} The sum converges because $f(x) = \mathcal{O}(1/x^2)$ by definition of the Schwartz space. It will turn out that in fact, $F \in \mathcal{S} \left( \mathbb{R} \right)$ as well. For this, we require the following lemma:
Lemma: Let $f_n : \mathbb{R} \longrightarrow \mathbb{C}$ be continuously differentiable for $n \in \mathbb{Z}_{> 0}$, and let $g : \mathbb{R} \longrightarrow \mathbb{C}$ be such that $f_n^\prime \longrightarrow g$ uniformly on every compact subset of $\mathbb{R}$. Suppose further that there is $c \in \mathbb{R}$ such that $f_n(0) \longrightarrow c$ as $n \to \infty$. Then, there is continuously differentiable $f : \mathbb{R} \longrightarrow \mathbb{C}$ with $f_n \longrightarrow f$ uniformly on every compact subset of $\mathbb{R}$, and $f^\prime = g$.
Proof. By fundamental theorem of calculus, \begin{align*} f_n(x) &= f_n(0) + \int_0^x f_n^\prime \left( t \right) \, dt \end{align*} Since $f_n^\prime \longrightarrow g$ uniformly on the compact interval $[0, x]$, we can take $n \to \infty$ and interchange the limit with the integral to obtain \begin{align*} \lim_{n \to \infty} f_n(x) &= c + \int_0^x g \left( t \right) \, dt \end{align*} which gives us the desired $f : \mathbb{R} \longrightarrow \mathbb{C}$. So far, we have only shown pointwise convergence $f_n \longrightarrow f$. To get uniform convergence on an arbitrary comapct subset $K \subseteq \mathbb{R}$, we use the bound \begin{align*} \left| f_n(x) - f(x) \right| &= \left| f_n(0) - c + \int_0^x f_n(t) - g(t) \, dt \right| \le \left| f_n(0) - c \right| + \int_0^x \left| f_n(t) - g(t) \right| \, dt \le \left| f_n(0) - c \right| + \int_0^M \left| f_n(t) - g(t) \right| \, dt \end{align*} where $M \ge 0$ is chosen such that $|x| \le M$ for all $x \in K$. Taking $n \to \infty$, the upper bound vanishes by uniform convergence of $f_n^\prime$ to $g$ which proves that $f_n \longrightarrow f$ uniformly on $K$. Finally, we need to show that $f$ is continuously differentiable with $f^\prime = g$. This follows because $g$ is the local uniform limit of continuous functions which makes $g$ continuous, and by the fundamental theorem of calculus, the derivative of $f$ is \begin{align*} \frac{d}{dx} f(x) &= \frac{d}{dx} \left( c + \int_0^x g(t) \, dt \right) = g(x) \end{align*} This completes the proof.
This gives the following consequence for periodisations of Schwartz functions:
Proposition: Let $f \in \mathcal{S} \left( \mathbb{R} \right)$, and let $F : \mathbb{R} \longrightarrow \mathbb{C}$ be \begin{align*} F(x) &= \sum_{n \in \mathbb{Z}} f\left( x + n \right) \end{align*} Then $F$ is smooth and $F(x + 1) = F(x)$ for all $x \in \mathbb{R}$.
Proof. Let $N \in \mathbb{Z}_{> 0}$. By condition of being a Schwartz function, there exists for every $k \in \mathbb{Z}_{\ge 0}$ some $M_k ge 0$ such that $$ \sup_{x \in \mathbb{R}} \left| x^2 f^{(k)} (x) \right| \le M_k $$ Then when $n \in \mathbb{Z}$ such that $|n| > N$, we have for all $x \in [-N, N]$ that $|x + n| > n - N$ so that $$ \left| f^{(k)} (x + n) \right| \le \frac{M_k}{(x + n)^2} \le \frac{M_k}{(n - N)^2} $$ So for each $k \in \mathbb{Z}_{\ge 0}$ we have the bound $|f^{(k)}| \le M_k / (n - N)^2$ on the interval $[-N, N]$. Then because $$ \sum_{|n| > N} \frac{M_k}{(n - N)^2} = 2 M_k \sum_{n = 1}^\infty \frac{1}{n^2} < \infty $$ we have by Weierstrass $M$-test the uniform convergence $$ \lim_{M \to \infty} \sup_{x \in [-N, N]} \left| \sum_{|n| > N} f^{(k)} (x + n) - \sum_{N < |n| \le M} f^{(k)} (x + n) \right| = 0 $$ It follows that we have uniform convergence $$ \lim_{M \to \infty} \sup_{x \in [-N, N]} \left| \sum_{n \in \mathbb{Z}} f^{(k)} (x + n) - \sum_{|n| \le M} f^{(k)} (x + n) \right| = 0 $$ Then because the sum $\sum_{n \in \mathbb{Z}} f^{(k)} (x + n)$ converges uniformly on every compact subset of $\mathbb{R}$, we have by the previous lemma that $$ \frac{d}{dx} \sum_{n \in \mathbb{Z}} f^{(k)} (x + n) = \sum_{n \in \mathbb{Z}} f^{(k + 1)}(x + n) $$ This proves that $F$ is infinitely differentiable as was claimed.
From this above result, it follows that we can equate the periodisation of a Schwartz function to its Fourier series. Recall also the Fourier transform defined for suitable functions $f : \mathbb{R} \longrightarrow \mathbb{C}$ by \begin{align*} \hat{f} \left( \xi \right) &= \int_\mathbb{R} e^{-2 \pi i x \xi} f(x) \, dx \end{align*} This will yield the Poisson summation formula:
Theorem (Poisson Summation Formula): Let $f \in \mathcal{S} \left( \mathbb{R} \right)$. Then, \begin{align*} \sum_{n \in \mathbb{Z}} f\left( x + n \right) &= \sum_{n \in \mathbb{Z}} \hat{f} \left( n \right) e^{2 \pi i n x} \end{align*}
Proof. Denote the infinite sum above on the left by $F$. The fact that the Fourier transform is well defined for a Schwartz function follows from the fact that $\mathcal{S} \left( \mathbb{R} \right) \subset L^1\left(\mathbb{R}\right)$. Taking the Fourier expansion of the periodisation gives \begin{align*} F(x) &= \sum_{n \in \mathbb{Z}} \left( \int_0^1 F(t) e^{- 2 \pi i n t} \, dt \right) e^{2 \pi i n x} \end{align*} By uniform convergence of $F$ on the compact interval $[0, 1]$ as proven in the previous proposition, we can interchange summation an integrals to compute the Fourier coefficients as \begin{align*} \int_0^1 F(t) e^{- 2 \pi i n t} \, dt &= \int_0^1 \lim_{N \to \infty} \sum_{m = -N}^N f(t + m) e^{- 2 \pi i n t} \, dt \\ &= \lim_{N \to \infty} \sum_{m = -N}^N \int_0^1 f(t + m) e^{- 2 \pi i n t} \, dt \\ &= \sum_{m \in \mathbb{Z}} \int_m^{m + 1} f(t) e^{- 2 \pi i n t} \, dt \\ &= \int_\mathbb{R} f(t) e^{- 2 \pi i n t} \, dt \\ &= \hat{f}(n) \end{align*} Thus, the Fourier expansion of $F$ is \begin{align*} \sum_{n \in \mathbb{Z}} f(x + n) &= \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{2 \pi i n x} \end{align*}