The image of $L^2(S^1)$ under the Fourier transform

Question

My lecturer stated following three statements:

1. The image of Fourier transform on Schwartz function on $\mathbb{R}$ is the Schwartz function on $\mathbb{R}.$

2. The image of $L^2(S^1)$ under the Fourier transform is $l^2(\mathbb{Z})$ and the image of $l^2(\mathbb{Z})$ under its Fourier transform is $L^2(S^1)$.

Here, $S^1$ indicates the unit circle. I don't need a rigorous proof for those statements. I just want a simple explanation why it holds. Especially for number $2$, the Fourier transform on $L^2(S^1)$ where $f\in L^2(S^1)$ is given $\hat{f}(k)=\int_0^1f(\theta)e^{-2\pi ik\theta}d\theta$. But why this tells us that the image is $l^2(\mathbb{Z})$?

Answer 1

For any orthonormal basis $(e_n)$ of a Hilbert space every element $x$ satisfies the property $(\langle x , e_n \rangle) \in \ell^{2}$ and any sequence $(a_n) \in \ell^{2}$ corresponds to some $x$ with $a_n=\langle x , e_n \rangle$ for all $n$.