Suppose that $ f \in L^1(\Bbb{R})\cap L^2(\Bbb{R})$ satisfies the following unity condition $$\sum_{k\in \Bbb{Z}} f(x-k) = 1\ \ \ \ , \forall x\in\Bbb{R}$$ Prove that $\hat{f}(0)=1$
Here $\hat{f}$is the fourier transform of $f$.
I tried writing $\int_{-\infty}^{\infty} f(t-k)e^{-i\omega t}dt = e^{-i\omega k}\hat{f}(\omega)$, which implies that $e^{i\omega k}\int_{-\infty}^{\infty} f(t-k)e^{-i\omega t}dt = \hat{f}(\omega)$
Therefore sub $\omega=0$ gives $\int_{-\infty}^{\infty} f(t-k)dt =\hat{f}(0)$
I am unsure how to proceed from there. Anyone can help me please?
We have $$\hat{f}\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx$$ then $$\hat{f}\left(0\right)=\int_{-\infty}^{\infty}f\left(x\right)dx=\sum_{k\in\mathbb{Z}}\int_{k}^{k+1}f\left(x\right)dx=\sum_{k\in\mathbb{Z}}\int_{0}^{1}f\left(x-k\right)dx=\int_{0}^{1}\sum_{k\in\mathbb{Z}}f\left(x-k\right)dx=1.$$