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It's a known proof, but here's what I've got so far. Let $f \in L^2_\mathbb{C}[-\pi,\pi]$, with $f(x)=x$, and take $e_n(x) = e^{inx}/\sqrt{2\pi}, n \in \mathbb{Z}$ which is an orthonormal sequence in the Hilbert space $L^2_\mathbb{C}[-\pi,\pi]$. Note that the inproduct is defined as $(f,g) = \int^\pi_{-\pi}f\overline{g}\,dx$.

Then by Parseval's theorem we have that: $$ \lVert x \rVert^2 = \sum_{n=1}^\infty|(x,e_n)|^2$$ for all $x \in L^2_\mathbb{C}[-\pi,\pi]$. We have that $ \lVert x \rVert^2 = \int^\pi_{-\pi}x^2 \, dx = \frac{2}{3}\pi^3$. Then calculating:

\begin{align*} (x,e_n) &= \frac{1}{\sqrt{2\pi}}\int^\pi_{-\pi}xe^{-inx}\,dx \\ &= \frac{1}{\sqrt{2\pi}}\left(\frac{ix}{n}e^{-inx} \right)_{x \in [-\pi,\pi]} \\ &= \frac{1}{\sqrt{2\pi}} \frac{2\pi i}{n}e^{-in\pi} = \frac{\sqrt{2\pi}i}{n}e^{in\pi} \end{align*} This means that $|(x,e_n)|^2 = \frac{2\pi}{n^2}$, so from the first equality we now have: \begin{align*} \lVert x \rVert^2 &= \sum_{n=1}^\infty|(x,e_n)|^2 \\ \frac{2}{3}\pi^3 &= 2\pi\sum_{n=1}^\infty \frac{1}{n^2} \\ \frac{\pi^2}{3} &= \sum_{n=1}^\infty \frac{1}{n^2}. \end{align*}

This is obviously wrong, but I'm trying to pinpoint where. I feel it has something to do with my books form of Parseval's theorem, which takes the sum from 1 to $\infty$, and not $-\infty$ to $\infty$, which causes the missing factor of 2. Can't see how this would remedied by anything. Any ideas?

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It would be remedied by using the correct form of Parseval's theorem, where the sum is over $\mathbb{Z}$, as you noticed. –  Ragib Zaman Jun 15 '12 at 10:17

2 Answers 2

$\{e_n:n\in\mathbb{Z}\}$ is an orthonormal sequence of the Hilbert space, but not $\{e_n:n\in\mathbb{Z},n>0\}$.

Then Parseval's theorem should be:

$\displaystyle\| x \|^2 = \sum_{n=-\infty}^{+\infty}|(x,e_n)|^2$

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Your basis is wrong, you have to account for all integer powers of $e^{ix}$. This then creates the extra $2$ that you require by symmetry of the squares (you will be summing reciprocal squares of ALL non-zero integers, this being the same as double the sum for just positive integers).

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