Proving that the Fourier Basis is complete for C(R/$2*\pi$ , C) with $L^2$ norm Let $H$ be the inner product space = $\{f: \mathbb{R} \to \mathbb{C} \mid f \text{ is continuous and has period }2 \pi\}$ where the inner product is:
$$\langle f,g \rangle = \int_{0}^{2\pi}f(t)\overline{g(t)} dt $$ 
How do I prove that for $n \in \mathbb{Z}$ and $e_n(t)=\dfrac{1}{\sqrt{2\cdot\pi}} e^{i n  t}$, $(e_n)_{n \in N}$ is a basis of $H$ (that it is orthonormal is easy).

This is probably not the most elegant way, but this method generalizes to showing the unitarity of the Fourier transform. The essential part of this proof is showing that the map $L^2(\mathbb{R}/2\pi \mathbb{Z}) \rightarrow \ell^2(\mathbb{Z})$ given by taking the Fourier coefficients from the discrete Fourier transform is an isometry, since the discrete Fourier transform is essentially the orthogonal projection onto the closure of the span of the functions $e^{inx}$. Then, by basic Hilbert space theory, it would follow that $\{e^{inx}\}$ is a maximal orthogonal set.
To start, let $\{a_n\} \subset \ell^2(\mathbb{Z})$ be the sequence of Fourier coefficients, a.k.a. $a_n = \langle f, e^{inx}/2\pi \rangle$. Then
\begin{align*}
\sum_{-n}^n |a_n|^2 & = \sum_{-n}^n \frac{1}{2 \pi} \int_0^{2\pi} \int_0^{2\pi} f(x) \overline{f(y)} e^{in(x-y)} \, dx \, dy \\
& = \frac{1}{2 \pi} \int_0^{2\pi} \int_0^{2\pi} f(x) \overline{f(y)} D_n(x-y) \, dx \, dy,
\end{align*}
where $D_n$ is the Dirichlet kernel given by $D_n(t) = \sum_{-n}^n e^{int}$. Now, we can prove several basic facts about this Dirichlet kernel:


*

*It integrates to $2\pi$.

*When $t \neq 0$, $D_n(t) \rightarrow 0$ as $n \rightarrow \infty$.


Therefore the Dirichlet kernel behaves as an approximate identity times $2\pi$, so $D_n(x-y)$ behaves like a Dirac mass about $x = y$. Substituting this in (one of course needs to use an appropriate theorem to justify integrating a limit), we have precisely that
$$ \sum_{-n}^n |a_n|^2 \rightarrow \|f\|_{L^2}^2, \quad n \rightarrow \infty$$
