One can start by considering the vector space $V$ of all square-integrable functions on $[-L,L]$ -- this gives us a vector space with an inner product. Specifically, we're interested in the subspace $V_n$ that is the span of complex exponentials upto $n$ and $-n$.Then given a vector $f$ in $V$, we can ask for its

**projection**$f_n$ onto $V_n$.

As the complex exponentials are already orthonormal, it is easy to calculate this projection in their basis:

\[\begin{gathered}

{a_k} = \left\langle {f,\frac{1}{\sqrt{2L}}{e^{2\pi i\;nx/L}}} \right\rangle = \int\limits_{ - L}^L {f(x)\frac{e^{ - 2\pi i\;kx/L}}{\sqrt{2L}}dx} \hfill \\

{f_n}(x) = \sum\limits_{|k|\le n} {{a_k}\frac{e^{2\pi i\;kx/L}}{\sqrt{2L}}} \hfill \\

\end{gathered} \]

Notably this implies by Cauchy-Schwarz that:

\[{\left| f \right|^2} \geqslant \sum\limits_{|k| \leqslant n} {{{\left| {{a_k}} \right|}^2}} \]

This really just is Cauchy-Schwarz, and is known as

**Bessel's inequality**. If we can show that the Fourier series approaches $f$, i.e. that $\left\|f-f_n\right\|\to 0$, then it would be obvious that

\[{\left| f \right|^2} = \sum\limits_{|k| \in \mathbb{Z}} {{{\left| {{a_k}} \right|}^2}} \]

Which is just the Pythagoras theorem, and is known as

**Parseval's theorem**. Obviously, these theorems exist in the general theory of Hilbert spaces.

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