Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis of $L^2(S^1)$ with Haar measure

Question

Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis.

1. We show that this is an orthonormal set.

All we need to show is that $\int_{S^1} z^n d\mu=0$ since $\mu$ is a Haar measure $\int_{S^1} z^n d\mu=\int_{S^1} (az)^n d\mu=a^n\int_{S^1} z^n d\mu$, but if $a$ is not a root of unity, this implies the integral is $0$. Thus the set is orthonormal.

2. We show the the span is dense in $C(S^1)$ which in turn is dense in $L^2(S^1)$ which proves it is an orthonormal basis.

The Span is clearly an algebra so all we have to show is that it separates points. But that is fairly obvious. $x\not = y$ then take the identity map.

Is this correct?

Answer 1

You missed another condition. You need the fact that the span contains the complex conjugate of each of its elements. For this note that the conjugate of $z^{n}$ is $\frac 1 {z^{n}}=z^{-n}$ when $|z|=1$.