I never really bothered to ask this question and now it seems silly...but why do we always seem to define function spaces $X(U)$ (e.g. $L^2(U), BV(U)$ for open sets $U\subset\Bbb{R}^n$? What breaks if we consider closed $U$? For example, $L^2([0,1]^2)$ seems perfectly natural to me.

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    $\begingroup$ $[0,1]$ appears frequently? In any case, wild functions prefer open spaces... $\endgroup$ – copper.hat Apr 8 '14 at 6:33
  • $\begingroup$ @copper.hat do you maybe have an example of a function/property that only make sense for open sets? (I would imagine some kind of topoligist's sine curve...?) $\endgroup$ – icurays1 Apr 8 '14 at 6:35
  • $\begingroup$ I don't really have any good examples. Maybe the space of invertible matrices? $\endgroup$ – copper.hat Apr 8 '14 at 6:39
  • $\begingroup$ I always thought it is because finite sets are always closed under the usual topologies and in such sets differentiability is problematic. $\endgroup$ – Git Gud Apr 8 '14 at 6:53

First, there is a difference between $C([0,1])$ and $C((0,1))$ - the spaces of functions that are continuous on the closed and open interval, respectively. There is no difference between $L^2([0,1])$ and $L^2((0,1))$ though.

If you define function spaces on open sets then every point of the set is an interior point, which is convenient if you want to work with differentiability.

  • $\begingroup$ Okay, the part about differentiability makes perfect sense...not sure why I didn't think about that. $\endgroup$ – icurays1 Apr 8 '14 at 16:17

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