Given $f\in L^1(\mathbb{R})$ such that $\widehat{f} \in C_c^\infty(\mathbb{R})$, let's define for $\lambda > 0$ the function $$\psi_\lambda(t) = \sum_{-\infty}^\infty f \left(\frac{t-2\pi n}{\lambda}\right)$$ then clearly $\psi_\lambda$ is $2\pi$-periodic and belongs to $L^1(\mathbb{T})$ where $\mathbb{T} = \mathbb{R}/2\pi \mathbb{Z}$. The question is, the author Katznelson in "Introduction to Harmonic Analysis" claim that (page 149)
"$\widehat{f} \in C_c^\infty(\mathbb{R})$" implies that for any $\varepsilon > 0$, there exists $\lambda>0$ small enough so that $$ \sup_{t\in \mathbb{T}} |\psi_\lambda(t)| < \sup_{x\in \mathbb{R}} |f(x)| + \varepsilon$$ Can I anyone explain it to me rigorously?
I think I got it, but comments are welcome.
Since $\widehat{f}\in \mathcal{S}(\widehat{\mathbb{R}})$ we have $f\in \mathcal{S}(\mathbb{R})$ as well, thus \begin{equation*} \sup_{x\in \mathbb{R}} (2\pi + |x|)^2 |f(x)| \leq C \qquad\Longrightarrow\qquad |f(x)| \leq \frac{C}{(2\pi+|x|)^2} \quad\text{for all}\quad x\in \mathbb{R}. \end{equation*} For $t\in \mathbb{T}\sim [0,2\pi)$ and $n\neq 0$ we have \begin{equation*} \left|f\left(\frac{t-2\pi n}{\lambda}\right)\right| \leq \frac{C\lambda^2}{\big(2\pi + |t-2\pi n|\big)^2} \leq \frac{C\lambda^2}{\big(2\pi + 2\pi |n| - t\big)^2} \leq \frac{C\lambda^2}{4\pi^2 |n|^2}. \end{equation*} Hence \begin{equation*} |\psi_\lambda(t)| \leq \left|f\left(\frac{t}{\lambda}\right)\right| + \sum_{n\neq 0} \left|f\left(\frac{t-2\pi n}{\lambda}\right)\right| \leq \sup_{x\in \mathbb{R}} |f(x)| + \frac{C\lambda^2}{4\pi^2} \sum_{n\neq 0}\frac{1}{n^2} = \sup_{x\in \mathbb{R}} |f(x)| + \frac{C\lambda^2}{24} \end{equation*} and thus the result follows when we choose $\lambda$ small enough.