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?


1 Answer 1


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 $$ \sum_{k=-\infty}^{+\infty} f(x+2kL)=\frac{\sqrt{2\pi}}{2L} \sum_{n=-\infty}^{+\infty} \hat{f}\left(\frac{n\pi}{L}\right)e^{in\pi x/L}. \tag{5} $$


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