Modified trigonometric system

Question

We know that the trigonometric system $\left\{e_{n}= e^{2 \pi i n x} \right\}_{n \in \mathbb{Z}}$ is complete in $L^{2}[0,1]$, using this fact, is there a smart way to show $ \mathcal{E}_{b}=\left\{e^{2 \pi i b n x}\right\}_{n \in \mathbb{Z}} $ is complete in $ L^{2}\left[0, b^{-1}\right] $ for $b>0$ a fixed scalar?

Answer 1

Let $d$ be the dilation $d:[0,1/b] \rightarrow [0,1]:x \mapsto bx$. Then observe $$ e_n \circ d= \mathcal{E}_n $$ and $$ e_n = \mathcal{E}_n \circ d^{-1} \text{.} $$

Try using this on $f \in L^1[0,1/b]$ as $f \circ d^{-1} \in L^1[0,1]$, so from $$ f \circ d^{-1} = \sum_i a_i e_i \text{,} $$ it is immediate that $$ f = \sum_i a_i e_i \circ d = \sum_i a_i \mathcal{E}_i \text{.} $$ That is, push $f$ forward from $[0,1/b]$ to $[0,1]$ using $d$, apply completeness there, then pull back using $d^{-1}$.