Why, if taking $\lim_{N \to \infty} \sum_{-N}^{N} \hat{f}(n) e^{2\pi i nx}$ allowed in $L^2$ (why is it not order dependent - where I can see this)? And - does this hold for any Hilbert space?
In particular, why in the $L^2$ norm the order doesn't matter, but in $| \cdot |$ (point-wise) it does? (which I suppose answers my second question in that if the space is equipped with the $| \cdot |$ norm then the order does matter? Or am I confusing things up?).
For infinite series in $\mathbf R$ or $\mathbf C$ (or $\mathbf R^n$), the order of summation does not matter if the series is absolutely convergent (the converse is true too).
Lebesgue integrability is an “absolute” integral: a measurable function $f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable (this is false for non-measurable functions: consider a function $f$ with value $1$ on a non-measurable subset of $[0,1]$ and $-1$ on its complement in $[0,1]$: $f$ is not Lebesgue integrable on $[0,1]$, but $|f|$ is). And a measurable function $f$ is in $L^p$ if and only if $|f|$ is in $L^p$.
In the case $p=2$, a Fourier series with coefficients $c_n$ is in $L^2([0,1])$ if and only if $\sum |c_n|^2$ converges, so absolute convergence of a Fourier series is built in to the meaning of $L^2$-convergence (but not pointwise convergence) of Fourier series.
