The inverse of the Fourier transform in $L^2[0,2\pi]$

Question

Let $x\in l^2(\mathbb{Z},\mathbb{C})$ and we define: $\mathcal{(F^*x)}(t)=\sum_{n\in \mathbb{Z}}[x_ne^{int}]$

And I want to prove that $\mathcal{F^*x}\in L^2[0,2\pi]$.

My idea:

$$\forall \epsilon>0 \, ,\,\exists N \in \mathbb{N }\,, \,|\sum_{|n|>N}(x_ne^{int})|\leq \epsilon.$$ Then $$|\sum_{n\in \mathbb{Z}}(x_ne^{int})|^2\leq (\sum_{|n|<N}|x_ne^{int}|)^2+\epsilon$$

then we use the inequality : $$|\sum_{|n|<N}x_n.e^{int}|^2<N^2(\sum_{|n|<N}|x_n|^2)$$ then we integrate on $[0,2\pi]$

Answer 1

It's simpler than that. The complex exponentials are orthogonal in $L^2([0, 2\pi])$.

$$\begin{split} \left|\sum_{|n|\leq p}x_ne^{int}\right|^2 &= \sum_{|n|\leq p}\left|x_n\right|^2 + \sum_{|k|\leq p\\|l|\leq p \\ k\neq l} x_kx_l^*e^{i(k-l)t} \end{split}$$ $$\int_0^{2\pi}\left|\sum_{|n|\leq p}x_ne^{int}\right|^2dt=2\pi \sum_{|n|\leq p}\left|x_n\right|^2 +\sum_{|k|\leq p\\|l|\leq p \\ k\neq l} x_kx_l^*\underbrace{\int_0^{2\pi}e^{i(k-l)t}dt}_{=0}=2\pi \sum_{|n|\leq p}\left|x_n\right|^2$$