Showing $\{e^{inx}\}$ is an orthonormal basis in $L^2(-\pi,\pi)$

Question

I want to show $\{e^{inx}\}$ is an orthonormal basis in $L^2(-\pi,\pi).$ I first showed $\{e^{inx}\}$ is an orthonormal set. Let $\varphi_n(x)=e^{inx}.$ Then \begin{equation*} \varphi_n(x)=\cos(nx)+i\sin(nx). \end{equation*} Hence \begin{equation*} \begin{split} \langle\varphi_m,\varphi_n\rangle&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{imx}\overline{e^{inx}}dx\\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(m-n)x}dx\\ &=\begin{cases}1&\text{ when } m=n,\\0 &\text{ when }m\neq n. \end{cases} \end{split} \end{equation*} This shows $\{e^{inx}\}$ is an orthonormal set. But how can I show $\{e^{inx}\}$ spans $L^2(-\pi,\pi)?$ In order to show that, I need to show for arbitrary $f\in L^2(-\pi,\pi)$, $f=\sum_{n=1}^{\infty}\langle f,\varphi_n\rangle\varphi_n.$ I know the concept, but I am not sure how to apply it in this case.

Answer 1

There are various approaches to this. For example, the span of the $e^{inx}$ is uniformly dense in the continuous $2\pi$-periodic functions by Stone-Weierstrass. Therefore this span in $L^2$-dense in the set of continuous functions on $[-\pi,\pi]$ with $f(\pi)=f(-\pi)$. That is $L^2$-dense in the set of all continuous functions on $[-\pi,\pi]$, which in turn is $L^2$-dense in $L^2(-\pi,\pi)$.