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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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
1 Answers
There is no difference between $L^2([−\pi,\pi))$, $L^2((−\pi,\pi))$ and $L^2([−\pi,\pi])$, because a singleton is a null set. My personal preference is for the latter notation.
But when discussing continuous functions, we pay attention even to single points. The reason to use $[-\pi,\pi)$ is that this interval is in bijection with the circle $S^1$ via $x\mapsto e^{ix}$. Bringing the topology of $S^1$ to $[-\pi,\pi)$, we discover that it is just like the standard topology, with one exception: when a sequence tried to converge to the missing endpoint $\pi$, it converges to $-\pi$ instead. Having made a note of this, we can define $C([-\pi,\pi))$ as the set of functions that are continuous in this circle topology. This is a different space from the space of continuous functions on unmodified interval, $C([-\pi,\pi])$. The latter space contains $f(x)=x$ which is not in $C([-\pi,\pi))$.