Let $L: [a, a + T] \rightarrow [-\pi, \pi]$, $L(x) := - \pi + \frac{2\pi}{T}(x-a)$.
Let $\left \{ \phi_n \right \}_{n \in \mathbb{N}} \subset L^2([-\pi, \pi])$ be an orthonormal set.
Show that $\left \{ \sqrt{\frac{2\pi}{T}} (\phi_n \circ L) \right \}_{n \in \mathbb{N}}$ is an orthonormal set in $L^2([a, a + T])$.
I managed to prove that this set is orthogonal, but I have some troubles showing that a norm of a fucntiontion from this set is $1$:
$\left \| \sqrt{\frac{2\pi}{T}} (\phi_n \circ L) \right \|^2$
$= \left \langle \sqrt{\frac{2\pi}{T}} (\phi_n \circ L), \sqrt{\frac{2\pi}{T}} (\phi_n \circ L) \right \rangle$
$= \frac{2\pi}{T} \left \langle \phi_n \circ L, \phi_n \circ L\right \rangle =$
$ = \frac{2\pi}{T} \left \langle \phi_n(- \pi + \frac{2\pi}{T}(x-a)) , \phi_n(- \pi + \frac{2\pi}{T}(x-a)) \right \rangle$
But this $\{\phi_n \}$ is orthonormal, this inner product would equal one, implying that $\left \| \sqrt{\frac{2\pi}{T}} (\phi_n \circ L) \right \| = \sqrt{\frac{2\pi}{T}}$, which contradicts the othonormality of the set.
What am I missing? How should I calculate the norm of this function composition?
Edit: Okay, I got this correctly based on this answer: Proving that a set $\{\psi_n(x)\}_1^\infty = \{\sqrt{c}\;\phi_n(cx+d)\}_1^\infty$ is an orthonormal basis. However, I'm still curious why my reasoning is incorrect.