Sum of continuous $L^{1}$ function over the integers. Let $f$ be a continuous $L^1$ function defined on $\mathbb{R}$, such that $\hat{f}(k) = 0$ for $|k| > 1/2$, where $\hat{f}$ is the Fourier transform. Is it true that $\sum_{k = -\infty}^{\infty}|f(-k)| < \infty$?
I know that $\int_{[0,1]} \sum_{k = -\infty}^{\infty}|f(x-k)| = \sum_{k = -\infty}^{\infty} \int_{[k,k+1]}|f(x)| = \int |f(x)| < \infty$, from which it follows that $\sum_{k = -\infty}^{\infty}|f(x-k)| < \infty$ almost everywhere. Is there someway to use the continuity of $f$ to conclude that this is true for all $x$ (and in particular $x=0$)?
 A: This is indeed true. The intuition is that a bandlimited function,
in your case with ${\rm supp}\left(\smash{{\widehat{f}}}\right)\subset\left[-\frac{1}{2},\frac{1}{2}\right]$
is essentially constant on intervals of length $\left(\frac{1}{2}\right)^{-1}=2$.
To make this precise, we need some theory of the Fourier transform
and of Schwartz functions. We choose a test function $\varphi\in C_{c}^{\infty}\left(\mathbb{R}\right)\subset\mathcal{S}\left(\mathbb{R}\right)$
with $\varphi|_{\left[-\frac{1}{2},\frac{1}{2}\right]}\equiv1$. This
yields $\psi:=\mathcal{F}^{-1}\varphi\in\mathcal{S}\left(\mathbb{R}\right)$.
By assumption on (the Fourier transform of) $f$, we have
$$
\widehat{f}=\widehat{f}\cdot\varphi.
$$
By the convolution theorem, this means 
$$
f=\mathcal{F}^{-1}\widehat{f}=\mathcal{F}^{-1}\left(\widehat{f}\cdot\varphi\right)=\left(\mathcal{F}^{-1}\widehat{f}\right)\ast\left(\mathcal{F}^{-1}\varphi\right)=f\ast\psi.
$$
By the assumed continuity of $f$, we have $f=\mathcal{F}^{-1}\widehat{f}$
with pointwise equality (not only almost everywhere).
Define, for arbitrary $R>0$
$$
M\psi\left(x\right):=\sup_{\left|y-x\right|<R}\left|\psi\left(y\right)\right|,
$$
and analogously $Mf\left(x\right):=\sup_{\left|y-x\right|<R}\left|f\left(y\right)\right|$.
Observe that for $\left|y-x\right|<R$,
$$
1+\left|x\right|\leq1+\left|x-y\right|+\left|y\right|\leq1+R+\left|y\right|\leq\left(1+R\right)\cdot\left(1+\left|y\right|\right)
$$
and hence (since $\psi$ is a Schwartz function)
$$
\left|\psi\left(y\right)\right|\leq C_{N,\psi}\cdot\left(1+\left|y\right|\right)^{-N}\leq C_{N,\psi}\cdot\left(1+R\right)^{N}\cdot\left(1+\left|x\right|\right)^{-N}=:C_{N,R,\psi}\cdot\left(1+\left|x\right|\right)^{-N},\qquad\left(\dagger\right)
$$
where $N\in\mathbb{N}$ is arbitrary.
We can now estimate for $x\in\mathbb{R}$ arbitrary and $\left|y-x\right|<R$:
\begin{eqnarray*}
\left|f\left(y\right)\right| & = & \left|\left(f\ast\psi\right)\left(y\right)\right|\\
 & = & \left|\int_{\mathbb{R}}f\left(z\right)\cdot\psi\left(y-z\right)\,{\rm d}z\right|\\
 & \leq & \int_{\mathbb{R}}\left|f\left(z\right)\right|\cdot\sup_{\left|w-x\right|<R}\left|\psi\left(w-z\right)\right|\,{\rm d}z\\
 & = & \int_{\mathbb{R}}\left|f\left(z\right)\right|\cdot\sup_{\left|v-\left(x-z\right)\right|<R}\left|\psi\left(v\right)\right|\,{\rm d}z\\
 & = & \int_{\mathbb{R}}\left|f\left(z\right)\right|\cdot\left(M\psi\right)\left(x-z\right)\,{\rm d}z\\
 & = & \left(\left|f\right|\ast M\psi\right)\left(x\right).
\end{eqnarray*}
As this holds for all $\left|y-x\right|<R$, we arrive at
$$
Mf\left(x\right)\leq\left(\left|f\right|\ast M\psi\right)\left(x\right).
$$
But the estimate $\left(\dagger\right)$ above yields (for $N\geq2$)
that $M\psi\in L^{1}\left(\mathbb{R}\right)$ and hence $\left|f\right|\ast M\psi\in L^{1}\left(\mathbb{R}\right)$,
which also implies $Mf\in L^{1}\left(\mathbb{R}\right)$.
As you noted yourself, this implies that there is some $x_{0}\in\left(-\frac{R}{2},\frac{R}{2}\right)$
with
$$
\sum_{k=-\infty}^{\infty}Mf\left(x_{0}-k\right)<\infty,
$$
this actually holds for almost all $x_{0}\in\mathbb{R}$.
But for arbitrary $x\in\left(-\frac{R}{2},\frac{R}{2}\right)$ and
$k\in\mathbb{Z}$, we have $\left|\left(x-k\right)-\left(x_{0}-k\right)\right|\leq\left|x\right|+\left|x_{0}\right|<R$
and hence
$$
Mf\left(x_{0}-k\right)=\sup_{\left|y-\left(x_{0}-k\right)\right|<R}\left|f\left(y\right)\right|\geq\left|f\left(x-k\right)\right|.
$$
Thus, we finally arrive at
$$
\sum_{k=-\infty}^{\infty}\left|f\left(x-k\right)\right|\leq\sum_{k=-\infty}^{\infty}Mf\left(x_{0}-k\right)<\infty,
$$
which even shows uniform convergence of the series on $\left(-\frac{R}{2},\frac{R}{2}\right)$.
As $R>0$ was arbitrary, this establishes everything you wished for.
Disclaimer: One can probably find a simpler proof :)
