Showing a sequence is not Schauder basis

I'm surprised no one answered this. 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)