[proofplan] Fej\'er's theorem gives uniform density of trigonometric polynomials in $C(\mathbb{T})$. Uniform convergence implies $L^2$ convergence on a finite measure space, so trigonometric polynomials are dense in a dense subspace of $L^2(\mathbb{T})$. A two-step density argument (approximate $f \in L^2$ by continuous $g$, then $g$ by a trigonometric polynomial $p$) shows the span of the trigonometric system is dense in $L^2(\mathbb{T})$, which is equivalent to completeness. [/proofplan] [step:Establish density of trigonometric polynomials in $C(\mathbb{T})$] By [Fej\'er's Theorem](/theorems/584), for every $f \in C(\mathbb{T})$ and $\varepsilon > 0$, there exists $N$ such that $\|\sigma_N f - f\|_{L^\infty} < \varepsilon$. Since $\sigma_N f$ is a trigonometric polynomial, trigonometric polynomials are uniformly dense in $C(\mathbb{T})$. [/step] [step:Extend density from $C(\mathbb{T})$ to $L^2(\mathbb{T})$] For any $f \in L^2(\mathbb{T})$ and $\varepsilon > 0$: Choose $g \in C(\mathbb{T})$ with $\|f - g\|_{L^2} < \varepsilon/2$ (continuous functions are dense in $L^2$ on a finite measure space). By the previous step, choose a trigonometric polynomial $p$ with $\|g - p\|_{L^\infty} < \varepsilon/(2\sqrt{2\pi})$. Then: \begin{align*} \|g - p\|_{L^2} \leq \sqrt{2\pi}\,\|g - p\|_{L^\infty} < \varepsilon/2. \end{align*} By the triangle inequality, $\|f - p\|_{L^2} < \varepsilon$. Since $p$ lies in $\operatorname{span}\{e^{inx}\}_{n \in \mathbb{Z}}$, the span of the trigonometric system is dense in $L^2(\mathbb{T})$. [/step] [step:Conclude completeness from density of the span] By the [Characterisation of Complete Orthonormal Systems](/theorems/541), density of the span in $L^2(\mathbb{T})$ is equivalent to completeness. Therefore $\{e^{inx}/\sqrt{2\pi}\}_{n \in \mathbb{Z}}$ is a complete orthonormal system in $L^2(\mathbb{T})$. [/step]