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Let $f\in L^2(\Omega)$, where $\Omega \subset \mathbb{R}^2$ is the unit square $[0,1] \times [0,1]$. Let $x\in \Omega$.

Suppose I evaluate $f$ at points from some direction that approach $x$. Similarly, suppose I evaluate $f$ at points from a different direction that approach $x$.

Question: Will these two limits coincide almost everywhere? I can't think of a counter example, and I'm not sure how to go about a proof...maybe use density of continuous functions in $L^2$?

Edit: See the notes in the link below. The sentence directly above equation (2.6) in these notes is perhaps a better explanation of what I'm trying to establish (the $[\cdot]$ operator is defined in subsection 2.1).

http://www.cimpa-icpam.org/archivesecoles/20140204154528/dgfem.pdf

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    $\begingroup$ There is absolutely no reason why the limits should exists, let alone be equal... $\endgroup$
    – Tryss
    Aug 30, 2015 at 21:18
  • $\begingroup$ "Suppose I evaluate $f$" You cannot evaluate an element of $L^2$ pointwise and expect a well-defined result. Review the definition of $L^p$ spaces. Even approaching points through entire smooth curves does not make sense, because smooth curves still have zero Lebesgue $2$-measure. $\endgroup$
    – guest
    Aug 30, 2015 at 21:53
  • $\begingroup$ @guest Rudin in Real and Complex Analysis defines $L^p$ as a space of bona fide functions. Same thing here.: math.uh.edu/~dlabate/PagesfromAnalysis.pdf It's $L^p$ the metric space that requires equivalence classes. $\endgroup$
    – zhw.
    Aug 30, 2015 at 22:07
  • $\begingroup$ @zhw. fair enough, but since the question did mention "density of continuous functions in $L^2$" I think he has the metric in mind. $\endgroup$
    – guest
    Aug 30, 2015 at 22:23
  • $\begingroup$ Take the indicator function on the rationals. The limits don't exist at any point, in either direction. $\endgroup$
    – Alex R.
    Aug 31, 2015 at 1:48

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