Orthonormal basis of $L^2(T)$

Question

Why is $\{e_n\mid n\in\mathbb{Z}\}$ an orthonormal basis of $L^2(T)$, where $T=\{z\in\mathbb{C}\mid |z|=1\}$, $e_n(z)=z^n$, and $\int_T f(z)\,dz:=\int_0^1f(e^{2\pi i t})\,dt$?

My try:

If $n=m$, it is $$\langle e_n, e_n\rangle=\int_0^1e_n(e^{2\pi i t})\overline{e_n(e^{2\pi i t})}\,dt=\int_0^1e^{2\pi i tn}e^{-2\pi i tn}dt=\int_0^1 1\,dt=t\lvert_0^1=1. $$

If $n\neq m$, I'm stuck: $$ \langle e_n, e_m\rangle=\int_0^1e_n(e^{2\pi i t})\overline{e_m(e^{2\pi i t})}\,dt=\int_0^1e^{2\pi i tn}e^{-2\pi i tm}\,dt=\frac{e^{2\pi i(n-m)}}{2\pi i(n-m)}-\frac{1}{2\pi i(n-m)} $$ It is not always zero... how to continue/what is wrong here?

Answer 1

$$ \frac{e^{2\pi i(n-m)}}{2\pi i(n-m)}=\frac{1}{2\pi i(n-m)}. $$ Notice that $e^{2\pi i(n-m)}=\cos(2\pi (n-m))+i\sin(2\pi (n-m))=1+i0=1$.

Answer 2

The space $L^{2}[0,1]$ is a Hilbert space, and $$ \partial f = \frac{1}{i}\frac{d}{dt}f $$ is selfadjoint on the domain $\mathcal{D}(\partial)$ consisting of all absolutely continuous functions $f \in L^{2}$ for which $f' \in L^{2}$ with periodic conditions $f(0)=f(1)$. Because $\partial$ has only point spectrum, the normalized eigenfunctions of $\partial$ form a complete orthonormal basis of $L^{2}$. Those eigenfunctions are $\{ e^{2\pi inx} \}_{n=-\infty}^{\infty}$. Essentially, this is a generalization of the theorem that Hermitian matrices have a complete orthonormal basis of eigenvectors.

The orthogonality follows from the inner product symmetry of $\partial$: $$ (\partial f,g) = (f,\partial g),\;\;\; f,g\in\mathcal{D}(\partial). $$ The matrix argument proves this: $$ (2\pi n-2\pi m)(e_n,e_m)=(\partial e_n,e_m)-(e_n,\partial e_m)=0 \\ \implies (e_n,e_m) = 0 \mbox{ for } n \ne m. $$ The completeness of the eigenfunctions is a more difficult matter that follows from the spectral theorem for selfadjoint linear operators on a Hilbert space.