Local uniform convergence of the periodized Gaussian

Put $\alpha =\pi \mathrm{/}\tau >0$ using [F5] and $\tau >0$, and set $q=\exp (-\alpha )\in (0,1)$ by [F4]. Fix $R>0$ and choose an integer $n_0>R+1$ by [F2]. For $\mathrm{\mid }x\mathrm{\mid }\le R$ and $\mathrm{\mid }n\mathrm{\mid }\ge n_0$, one has $\mathrm{\mid }x+n\mathrm{\mid }\ge \mathrm{\mid }n\mathrm{\mid }-R>1$, hence $(x+n{)}^2\ge (\mathrm{\mid }n\mathrm{\mid }-R{)}^2\ge \mathrm{\mid }n\mathrm{\mid }-R$. Since $\exp$ is increasing, the $n$th nonnegative summand is at most $\exp (-\alpha (\mathrm{\mid }n\mathrm{\mid }-R))=\exp (\alpha R)q^{\mathrm{\mid }n\mathrm{\mid }}$ by [F6]. The two-sided geometric tail is finite and tends to zero uniformly in $x\in [-R,R]$ as its starting index tends to infinity, by [F3]. This proves absolute and uniform convergence on every compact interval.

Each finite partial sum is continuous by [F1] and [F4]. Fix $x_0\in \mathbb{R}$ and $\epsilon >0$, choose $R>\mathrm{\mid }{x}_0\mathrm{\mid }+1$, and choose a partial-sum index for which the uniform tail on $[-R,R]$ is below $\epsilon \mathrm{/}3$ by step 1.1. Continuity of that finite partial sum at $x_0$ gives a neighborhood where its change is below $\epsilon \mathrm{/}3$; the two tail values contribute less than $2\epsilon \mathrm{/}3$. Thus the full sum is continuous at $x_0$. Absolute convergence permits the reindexing $m=n+1$ in ${\varphi }_{\tau }(x+1)={\sum }_n\exp (-\alpha (x+1+n{)}^2)$, giving ${\varphi }_{\tau }(x+1)={\varphi }_{\tau }(x)$. Since $x_0$ was arbitrary, the sum is continuous and one-periodic.