# Show {$e^{2\pi i bnx}$} is a basis for $L^2[0,b^{-1}]$

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.

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