Proving that orthonormal set is an orthonormal basis If I know that the set of functions $\{\phi_n\}_1^\infty$ forms an orthonormal basis on $L^2(a,b)$ and the set $\{\psi_n\}_1^\infty$ is an orthonormal set on $L^2(\frac{a-d}{c}, \frac{b-d}{c})$, with $c>0, d\in \mathbb{R}$ and it is given that: 
$$\psi_n(x) = \sqrt{c}\;\phi_n(cx+d),$$
then how do I prove that the set $\{\psi_n\}_1^\infty$ is an orthonormal basis on $L^2(\frac{a-d}{c}, \frac{b-d}{c})$?
I am given that if $f\in L^2(a,b)$ and the set $\{\phi_n\}_1^\infty$ forms an orthonormal basis on $L^2(a,b)$, then 
$$||f||^2 = \sum_{n=1}^\infty\left| \langle f, \phi_n \rangle \right|^2= \sum_{n=1}^\infty\left| \int_a^bf(x)\overline{\phi_n(x)}\;dx \right|^2\;\;\;\;\text{(by Parseval's equation)}.$$
How do I prove in the case of $\{\psi_n\}_1^\infty$ that if $f\in L^2(\frac{a-d}{c}, \frac{b-d}{c})$, then  
$$||f||^2 = \sum_{n=1}^\infty\left| \langle f, \psi_n \rangle \right|^2= \sum_{n=1}^\infty\left| \int_{\frac{a-d}{c}}^{\frac{b-d}{c}}f(x)\overline{\psi_n(x)}\;dx \right|^2?$$
UPDATE: 
Should I use the following to prove this: 
Let $\{\textbf{u}_1, ..., \textbf{u}_k\}$ be an orthonormal set of $k$ vectors in $\mathbb{C}^k$. For any $\textbf{a} \in \mathbb{C}^k$ we have:
$$||\textbf{a}||^2 = |\langle \textbf{a}, \textbf{u}_1\rangle|^2 + \cdots |\langle \textbf{a}, \textbf{u}_k\rangle|^2.$$
Is it sufficient to use this theorem in the proof?

Hints:
You goal is to prove that  $\{\psi_n\}_1^\infty$ is a linearly independent and spanning family.
Why is it already linearly independent ?
For the spanning property, consider $f \in L^2(\frac{a-d}{c}, \frac{b-d}{c})$
What can you say about $f(cx+d)$ ?
