I've been wasting my entire morning trying to understand a simple, short paragraph from Stein-Weiss, Introduction to Fourier Analysis in Euclidean spaces and I've miserably failed so far. Could you please help me and shed some light?
The ingredients of the failure: $f \colon \mathbb R^n \to \mathbb R$ is a continuous function with sufficiently fast decay at infinity, i.e. $$ |f(y)| \le C(1+|y|)^{-n-\delta}, \qquad \text{ for some } \delta >0 $$ (say, for every $y$) and we assume that its Fourier transform has the same decay property. (A very easy case I've been considering is if $f$ is in the Schwartz space)
Then the authors introduce the periodised version of $f$, defined as $$\tag{*}\label{periodicas} x \mapsto \sum_{z \in \mathbb Z^n} f(x+z) $$ and they claim:
By comparison with $\sum_{z \in \mathbb Z^n}(1+|z|)^{-n-\delta}$, we see that \eqref{periodicas} is a uniformly convergent series whose terms are continuous functions.
Thus the function in \eqref{periodicas} would be shown to be continuous.
Now, there is no doubt that the terms in the sum are continuous functions, nor do I have any problems in remembering that uniform limit of continuous functions is continuous. The issue is the uniform convergence. How is that obtained? Some sort of $M$-test or comparison criterion? By assumption we have $$ |f(x+z)| \le \frac{C}{(1+|x+z|)^{n+\delta}} $$ but it is not quite clear to me how to manipulate the RHS and get rid of that $x$... any suggestions? Thanks!
You can use the inequality \begin{equation}\tag{1} 1+|x+z|\ge (1+|z|)(1+|x|)^{-1} \end{equation} to show $$ \sup_{x\in[0,1)^n}|f(x+z)| \le \sup_{x\in[0,1)^n}C(\frac{1+|x|}{1+|z|})^{n+\delta}\le C'\frac{1}{(1+|z|)^{n+\delta}}. $$ By the $M$-test, the series $\sum_{z\in\mathbb Z^n}f(x+z)$ is uniformly convergent in $x\in[0,1)^n$.
To prove $(1)$, use the triangle inequality: \begin{align*} 1+|z| &\le 1+|z-(-x)|+|\!-\!x|\\ &=1+|z+x|+|x|\\ &\le(1+|z+x|)(1+|x|). \end{align*}