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Let $b > 0$ be a fixed positive scalar. Show {$e^{2\pi i bnx}$} for $n \in \mathbb{Z}$ is a orthogonal (but not orthonormal) basis for $L^2[0,b^{-1}]$.

I was able to show it is orthogonal and not orthonormal. All is left is to show that it is a basis. I know that {$e^{2\pi ni x}$} is an orthonormal basis for $L^2[0,1]$, and I want to use this fact.

So for an $f \in L^2[0,b^{-1}]$, I hope to extend it to $L^2[0,1]$, and form a linear combination from the basis {$e^{2\pi ni x}$}. From there somehow make a change of variable back to $L^2[0,b^{-1}]$. I could be on the wrong track however.

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    $\begingroup$ Don’t extend $f$. Consider instead $g(x)=f(x/b)$ and prove $g \in L^2([0,1])$. $\endgroup$ – Mindlack Feb 12 at 23:35
  • $\begingroup$ Ah great, thanks! It kinda just falls into place after that. $\endgroup$ – HCS Feb 13 at 0:10

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