As part of an exercise, I was asked to show that given an orthonoraml basis $(\varphi_1,\varphi_2,\varphi_3,...)$ in $L_2[-\pi,\pi]$, we can construct an orthonormal basis $(\psi_1,\psi_2,\psi_3,...)$ in $L_2[a,b]$, by defining: $$\psi_n(x) := \sqrt{\frac{2\pi}{b-a}}\varphi_n\left(\frac{2\pi}{b-a}\left(x-\frac{a+b}{2}\right)\right)$$ It is easy to show that $\psi_1,\psi_2,\psi_3,...$ is orthonormal system (since $\int_a^b\psi_n\overline{\psi_m}=\int_{-\pi}^\pi\varphi_n\overline{\varphi_m}$), and also that it is complete (since $g\perp\varphi_n$ iff $f\perp\psi_n$, where $g(x)=f\left(\frac{b-a}{2\pi}x + \frac{a+b}{2}\right)$). To confirm the answer, I tried checking Parseval's identity for $(\psi_1,\psi_2,\psi_3,...)$:
Given $f\in L_2[a,b]$, define $g$ as before, and we get: $$ \|f\|^2_{\left[a,b\right]}=\int_a^b{|f(x)|^2}dx =\frac{2\pi}{b-a}\int_{-\pi}^\pi{|g(t)|^2}dt =\frac{2\pi\|g\|^2_{\left[-\pi,\pi\right]}}{b-a} =\frac{2\pi}{b-a}\sum_n\left|\left<g,\varphi_n\right>_{\left[-\pi,\pi\right]}\right|^2 $$ Next, we calculate: $$ \left<f,\psi_n\right>_{\left[a,b\right]} =\sqrt{\frac{2\pi}{b-a}}\int_a^b{f(x)\overline{\varphi_n}\left(\frac{2\pi}{b-a}\left(x-\frac{a+b}{2}\right)\right)}dx =\sqrt{\frac{b-a}{2\pi}}\int_{-\pi}^\pi{g(t)\overline{\varphi_n}(t)}dt $$ So we get: $ \left<f,\psi_n\right>_{\left[a,b\right]} =\sqrt{\frac{b-a}{2\pi}}\left<g,\varphi_n\right>_{\left[-\pi,\pi\right]} $, and thus: $\|f\|^2_{\left[a,b\right]}=\left(\frac{2\pi}{b-a}\right)^2\sum_n\left|\left<f,\psi_n\right>_{\left[a,b\right]}\right|^2$. So it seems as though Parseval identity's does not hold in general. Where did I make a mistake?
It's a simple calculation error. When substituting to convert $\int_a^b \lvert f(x)\rvert^2\,dx$ into the integral over $\lvert g(t)\rvert^2$, we have
$$x = \frac{b-a}{2\pi}t + \frac{a+b}{2},$$
and hence obtain
$$dx = \frac{b-a}{2\pi}\,dt.$$
So
$$\int_a^b\lvert f(x)\rvert^2\,dx = \int_{-\pi}^\pi\left\lvert f\left(\frac{b-a}{2\pi}t+\frac{a+b}{2}\right)\right\rvert^2\frac{b-a}{2\pi}\,dt = \frac{b-a}{2\pi}\int_{-\pi}^\pi \lvert g(t)\rvert^2\,dt.$$