Showing a sequence is not Schauder basis I'm tring to show that in $L^2[0,1]$, there are at least two distinct choices of coefficients  $\left(c_{n}\right)_{n \in \mathbb{Z}}$ such that for $0<b<1$
$$1=\sum_{n \in \mathbb{Z}} c_{n} e^{2 \pi i b n x}.$$
One seems obvious that $c_0=1$ and other $c_n$ are all $0$. How to determine other choices of $c_n$? Also I want to show this series converges in $L^2$-norm so $
\left\{e^{2 \pi i b n x}\right\}_{n \in \mathbb{Z}}
$is not a Schauder basis.
 A: Below, I use the notation $e_a(x) := e^{2\pi i ax}$ for any real number $a$.
We clearly have the characteristic function  $\chi_{[0,b]} \in L^2[0,1] $ for $b<1$. Expressing it as a Fourier series (equality is in $L^2[0,1]$ ), we get
$$\chi_{[0,b]} = \sum_{n \in \mathbb{Z}} c_n e_n  $$
where  $c_n = \langle \chi_{[0,b]} , e_n \rangle $. By applying a substitution , we have
\begin{align}
 \langle \chi_{[0,b]} , e_n \rangle & = \int_{0}^{b} e^{-2\pi i n x} \ dx  \\
& = \int_0^1 b e^{-2\pi i bn x} dx \\
& = b\langle 1 , e_{nb} \rangle   
\end{align}
For any $f \in L^2[0,1]$, we consider the operator $D_{b}(f)(x) := f(bx)$. Then we have $D_{b} \chi_{[0,b]} (x) = 1 $  (Since $\chi_{[0,b]}(bx) = 1$ on $[0,1]$.)
So applying $D_{b}$ to both sides of the Fourier series for $\chi_{[0,b]}$ gives us
$$1 = b\sum_{n \in \mathbb{Z}} \langle 1, e_{nb}\rangle e_{nb}   $$
So $c_n= b\langle 1, e_{nb} \rangle $. This choice of $c_n$ is clearly different from the one you observed.
This proof requires knowing that $D_b: L^2[0,1] \to L^2[0,1]$ is a continuous operator but this is straightforward to check.
(Small tip: When you're talking about $L^2$ convergence, get rid of the $x$ variable as to not confuse it with pointwise convergence)
