In our lecture we proved following version of the PSF:
Assume $f \in L^1(\mathbb{R})$ and $$ \exists C,\varepsilon>0: \quad |f(t)|+|\hat{f}(t)|\le C(1+|t|)^{-1-\varepsilon} \quad \forall t\in\mathbb{R}$$ then $f,\hat{f}$ are continuous (follows from inversion formula) and $$ \underbrace{\sum_{k\in\mathbb{Z}}\ f(t-k)}_{:=\ F(t)}=\sum_{k\in\mathbb{Z}} \ \underbrace{\hat{f}(k)}_{=\ \hat{F}(k)}e^{2\pi i nt} \quad \forall t\in\mathbb{T}$$
[We call the LHS the periodization $F(t)$ of $f$ and one can proof, that $F\in L^1(\mathbb{T})$ and $\hat{f}(k)=\hat{F}(k)$. (1)]
I wonder, why the assumption $f \in L^1(\mathbb{R})$ and $$ |\hat{f}(t)|\le C(1+|t|)^{-1-\varepsilon} \quad (2)$$ is not enough. My argument would be as follow:
- (2) guarantees the absolute convergence of the LHS, therefore (using (1)), we get the uniform convergence of the Fourier series of F against some continuous function h. $$\sum_{k\in\mathbb{Z}} \hat{f}(k)e^{2\pi i nt}=\sum_{k\in\mathbb{Z}} \hat{F}(k)e^{2\pi i nt}=h(t) \quad \forall t \in \mathbb{T}$$
- The Fourier coefficients of $h-F$ have to be zero, therefore we have $h=F$