I have the following problem I need to solve:
Suppose $\{\phi_n\}_1^\infty$ is an orthonormal basis in $L^2(a,b)$ (set of square-integrable functions on $[a,b]$). Suppose $c>0$ and $d\in \mathbb{R}$, and let $\psi_n(x) = c^{1/2}\phi_n(cx+d)$. Show that $\{\psi_n\}_1^\infty$ is an orthonormal basis for $L^2(\frac{a-d}{c}, \frac{b-d}{c})$.
I tried firstly to show that the dot product (as defined in my book for complex functions)
$$\langle\psi_n, \psi_m\rangle = \int_{\frac{a-d}{c}}^{\frac{b-d}{c}}\psi_n(x)\overline{\psi_m(x)}\;dx$$
equals $1$, when $n=m$ and equals $0$, when $n\neq m$. So:
$$\langle\psi_n, \psi_m\rangle = \int_{\frac{a-d}{c}}^{\frac{b-d}{c}}\psi_n(x)\overline{\psi_m(x)}\;dx = \int_{\frac{a-d}{c}}^{\frac{b-d}{c}}\sqrt{c}\;\phi_n(cx+d)\overline{\sqrt{c}\;\phi_m(cx+d)}\;dx$$
$$=c\int_{\frac{a-d}{c}}^{\frac{b-d}{c}}\phi_n(cx+d)\overline{\phi_m(cx+d)}\;dx = c\int_{a}^{b}\phi_n(x)\overline{\phi_m(x)}\;dx.$$
Hmm, I guess I did a mistake somewhere? I know that, because $\{\phi_n\}$ is an orthonormal basis in $L^2(a,b)$ then:
$$\int_{a}^{b}\phi_n(x)\overline{\phi_m(x)}\;dx = \left\{ \begin{array}{lr} 1 & : n=m\\ 0 & : n\neq m \end{array} \right.$$
but for the functions $\{\psi_n\}$ I got
$$\langle\psi_n, \psi_m\rangle =\left\{ \begin{array}{lr} c & : n=m\\ 0 & : n\neq m \end{array} \right.$$
Could someone give me a push to the right direction? =) Where's my mistake? :)
P.S.
here is some reference from my book:
When you make the change of variables $t=cx+d$ you get $dx=\frac{dt}{c}$. So you are missing the $\frac{1}{c}$ which results from the change of variables. Otherwise, everything is ok.