# Poisson formula with fourier transform

if $$f\in C^{1}(\mathbb{R})$$ is such that, exists $$K>0$$ with $$|f(x)|+|f'(x)|\leq K(1+x^2)^{-1}$$ for all $$x\in \mathbb{R}$$, then, for all $$L>0$$:

$$\sum_{k=-\infty}^{+\infty} f(x+2kL)=\frac{\sqrt{2\pi}}{2L}\sum_{n=-\infty}^{+\infty} \hat{f}(\frac{n\pi}{L})exp(\frac{in\pi x}{L})$$

I have been trying to test this taking into account the Fourier transform, but I have not been able to achieve it, could you help me?

Let's start by noting that $$F(x)=\sum_{k=-\infty}^{+\infty} f(x+2kL) \tag{1}$$ is a periodic function with period $$2L$$: $$F(x+2L)=\sum_{k=-\infty}^{+\infty} f(x+2L+2kL)=\sum_{k'=-\infty}^{+\infty} f(x+2k'L)=F(x). \tag{2}$$ Therefore, it can be represented as a Fourier series in the interval $$[-L,L]$$: $$F(x)=\sum_{n=-\infty}^{\infty}F_n\,e^{in\pi x/L}, \tag{3}$$ where \begin{align*} F_n&=\frac{1}{2L}\int_{-L}^{L}F(x)\,e^{-in\pi x/L}\,dx \\ &=\frac{1}{2L}\int_{-L}^{L}\sum_{k=-\infty}^{\infty}f(x+2kL)\, e^{-in\pi x/L}\,dx \\ &=\frac{1}{2L}\sum_{k=-\infty}^{\infty}\int_{(2k-1)L}^{(2k+1)L} f(y)\,e^{-in\pi(y-2kL)/L}\,dy \\ &=\frac{1}{2L}\int_{-\infty}^{\infty}f(y)\,e^{-in\pi y/L}\,dy =\frac{\sqrt{2\pi}}{2L}\hat{f}\left(\frac{n\pi}{L}\right), \tag{4} \end{align*} where $$\hat{f}$$ is the Fourier transform of $$f$$. Combining $$(1)$$, $$(3)$$ and $$(4)$$ we finally obtain $$\sum_{k=-\infty}^{+\infty} f(x+2kL)=\frac{\sqrt{2\pi}}{2L} \sum_{n=-\infty}^{+\infty} \hat{f}\left(\frac{n\pi}{L}\right)e^{in\pi x/L}. \tag{5}$$