An old exam question I'm practicing with:
Let $f \in L^1 (\mathbb{R})$ and assume that $\hat{f}$ (the Fourier transform of $f$) is supported on the interval $[-1/2, 1/2]$. Let $\operatorname{sinc} (x) = \frac{\sin(x)}{x}$. Prove that $$f(x) = \sum_{n \in \mathbb{Z}} f(n) \operatorname{sinc}(x-n).$$
The Fourier transform of a function $g$ is given by $\hat{g}(\omega) = \int_{-\infty}^\infty g(x) e^{-2 \pi i x \omega} dx.$
It's not clear to me why the fact that $\hat{f}$ is compactly supported is important.
Best Answer
You can start with the Fourier series $$\sum_k\hat{f}(y+k)= \sum_n \widehat{\hat{f}}(n)e^{2i \pi ny} = \sum_n f(n)e^{-2i \pi ny}$$ where $\widehat{\hat{f}}(n) = f(-n)$ is the Fourier inversion theorem.
That$f$ is $L^1$ means $\hat{f}$ is continuous and that it is supported on $[-1/2,1/2]$ means the LHS is well-defined, that the Fourier series converges in $L^2_{loc}$, that $f$ is continuous, and that in $L^2$ $$\hat{f}(y) =1_{y \in [-1/2,1/2]} \sum_k\hat{f}(y+k)=1_{y \in [-1/2,1/2]} \sum_n f(n)e^{-2i \pi ny}= \sum_n f(n)1_{y \in [-1/2,1/2]} e^{-2i \pi ny}$$ The obtained equality thus holds in $L^1$ and inverse Fourier transform of both side gives $$f(x) = \sum_n f(n)\frac{\sin(\pi (x-n))}{\pi (x-n)}$$