Why is it that in many texts the natural domain of choice is $L^2 ( [- \pi, \pi) ) $ as opposed to $L^2 ( [- \pi, \pi] ) $? I would to think of the space $L^2 ( [- \pi, \pi] ) $ as the completion of $C_{c}( [- \pi, \pi] )$, so that I can use the fact that trig functions have compact support on the interval $[- \pi,\pi]$, but many books use $[- \pi, \pi)$.
Interval type for Fourier Analysis on $L^2( [-\pi,\pi) ) $
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analysis
fourier-series
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2Probably because you are thinking of functions with period $2\pi$ and for such functions $f(-\pi)=f(\pi)$. Why can't you just think of $L^2([-\pi,\pi))$ as the completion of $C_c([-\pi,\pi))$? – 2012-03-12
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0@ShawnD because the trig functions have compact support on $[-\pi, \pi]$ when restricted to that interval, and $[-\pi, \pi)$ is not compact. And in which domain do you mean we think of the functions with period $2\pi$? – 2012-03-13
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0I don't understand comment about support of trig functions. Anyway, a Cauchy sequence of functions on $[-\pi,\pi]$ is still Cauchy when you consider restriction to $[-\pi,\pi)$. Functions of period $2\pi$ on $\mathbb{R}$ are specified by their value on $[-\pi,\pi)$. – 2012-03-13
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0Don't forget that $L^2$ is not a space of functions but rather consists of *classes* of functions modulo functions that are zero almost everywhere. The single point $\pi$ thus doesn't matter. If you want to think of $L^2$ as a completion you should really work on the (compact) circle *group* $S^1$ and angular measure and complete the space of continuous functions $C(S^1)$ with respect to the $L^2$ norm (the space $C(S^1)$ can be identified with the set of continuous functions $f: [-\pi,\pi] \to \mathbb{C}$ such that $f(-\pi) = f(\pi)$.) – 2012-03-13