Suppose $f : [a,b] \to [c,d]$ and $f'(x) > 0$ for $ x \in [a,b]$. Show that if $\{ \phi_n \}$ is an orthonormal basis for $L^2 (c,d)$, then $ \{ \phi_n \circ f \} $ is an orthonormal basis for $L^2_{f'} (a,b)$.
By the substitution rule $$ \int_{a}^{b} \phi_n(f(u)) \overline{\phi_m(f(u))}f'(u) \ du = \int_{c}^{d} \phi_n(x) \overline{\phi_m(x)} \ dx $$
which implies that $\{ \phi_n \circ f \}$ is orthonormal in $L^2_{f'}(a,b)$.
To show that it is a basis of $L^2_{f'}(a,b),$ I have to show that if $g \in L^2_{f'}(a,b)$, then $g =\sum_{n=1}^{\infty}\langle g, \phi_n' \rangle \phi_n'$, where $\phi_n' = \phi_n \circ f$. I would assume it's another substitution, but I'm not sure how to show this.
Consider any $h$ in $L^2(c,d)$. We note that for any $\epsilon > 0$, there exists an $N_0$ such that for $N > N_0$, $$ \newcommand{\ip}[1]{\left\langle #1 \right\rangle} \int_a^b \left|h - \sum_{n=1}^N \ip{h,\phi_n}\phi_n\right|^2\,dx \leq \epsilon $$ We then note that $h \circ f \in L^2_{f'}(a,b)$, and that $$ \int_c^d \left|h \circ f - \sum_{n=1}^N \ip{h \circ f,\phi_n'}\phi_n'\right|^2 f'(u)\,du =\\ \int_a^b \left|h - \sum_{n=1}^N \ip{h,\phi_n}\phi_n\right|^2\,dx \leq \epsilon $$ So, we note that every function of the form $h \circ f$ with $h$ in $L^2(c,d)$ is within the span of our new orthonormal basis.
Now, note that $f$ has a left-inverse $f^{-1}$ such that $f^{-1} \circ f = \operatorname{id}_{[a,b]}$. Note furthermore that if $g \in L^2_{f'}(a,b)$, then $g \circ f^{-1} \in L^2(c,d)$. Now, every $g$ can be written in the form $g = (g \circ f^{-1}) \circ f$, which means that it's within the span of $\{\phi_n'\}$.