how do i prove that $\check{\widehat{\left(f\right)}}=f$

Question

I have by definition, $$\check{\alpha}(x):=f(x)=\sum_{k=-\infty}^{+\infty}\alpha_{k}e^{ikx}, \ \ x \in \mathbb{R} .$$ where $\alpha=(\alpha_{k})_{k\in \mathbb{Z}}\in \mathcal{S}(\mathbb{Z}) $, $\mathcal{S}(\mathbb{Z})$ is the space of rapidly decreasing sequences such that $$\sum_{k=-\infty}^{+\infty} |\alpha_k|<+\infty,\ \ \sum_{k=-\infty}^{+\infty} |\alpha_k||k|^{n}<+\infty, \ \ n=1,2,...$$ I have managed to prove that $\widehat{(\check{\alpha})}=\hat{f}=\alpha$ \begin{eqnarray*} \int_{-\pi}^{\pi}f(x)e^{-i\xi x}dx&=&\int_{-\pi}^{\pi}\sum_{k=-\infty}^{+\infty}\alpha_{k}e^{ikx}e^{-i\xi x}dx\\ &\underbrace{=}_{\text{Con.uniforme}}&\sum_{k=-\infty}^{+\infty}\alpha_{k}\int_{-\pi}^{\pi}e^{ikx}e^{-i\xi x}dx\\ &=&\alpha_{\xi}\int_{-\pi}^{\pi}e^{(\xi-\xi)i x} dx+ \sum_{k\neq \xi}\alpha_{k}\int_{-\pi}^{\pi}e^{(k-\xi)ix} dx\\\\ &\underbrace{=}_{\text{Ortogonalidad (2.11)}}& \alpha_{\xi}2\pi+\sum_{k\neq \xi}\alpha_{k}(0)\\ \dfrac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-i\xi x}dx&=&\alpha_{\xi}\\ \hat{f}(\xi)&=&\underbrace{\alpha_{\xi} }_{\in \ \mathcal{S}(\mathbb{Z})} \end{eqnarray*} but I can't prove that $\check{\widehat{\left(f\right)}}=f$, where $f \in C_{per}^{\infty}([-\pi,\pi])$.