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Let $\Omega \subset \mathbb R^d$ be an open bounded set with Lipschitz boundary. Let us consider $\Omega_\delta = \Omega \cap \delta\mathbb Z^d$ for $\delta >0$. I want to say that the measure of $\Omega_\delta$ is bounded by a constant which is independent of $\delta$ and somehow depending only on the original $\Omega$.

How can this be proved? I guess it is a classical result in numerical analysis, but I can't find a reference.

My motivation is this: suppose that we prove a discrete Poincaré inequality in $\Omega_\delta$. Then I want to get out of it the "continuous" Poincaré inequality without the measure of the domain blowing up in the limit. Does it make sense?

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  • $\begingroup$ The set $\Omega$ is open where? And what measure do you mean? And you probably mean something different than “$\Omega\cap \delta\mathbb{Z}$”, since $ \delta\mathbb{Z}$ is a countable subset of $\mathbb R$. $\endgroup$ Commented Oct 16, 2022 at 10:50
  • $\begingroup$ @PietroMajer Yes, sorry. I meant $\Omega \subset \mathbb R^d$ and the typo is that the post should read $\mathbb Z^d$ $\endgroup$
    – Hiro
    Commented Oct 16, 2022 at 10:53
  • $\begingroup$ @PietroMajer And I suppose that the appropriate mesure would be the one that counts the points in $\Omega_\delta$ $\endgroup$
    – Hiro
    Commented Oct 16, 2022 at 10:54
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    $\begingroup$ It seems a bit odd to expect a bound independent of $\delta$, no? I would think that $\lvert \Omega \cap \delta \mathbf{Z}^d \rvert$ would grow like $\delta^{-d}$. $\endgroup$
    – Leo Moos
    Commented Oct 16, 2022 at 11:15
  • $\begingroup$ @LeoMoos My motivation is this: suppose that we prove a discrete Poincaré inequality in $\Omega_\delta$. Then I want to get out of it the "continuous" Poincaré inequality without the measure of the domain blowing up in the limit. Does it make sense? $\endgroup$
    – Hiro
    Commented Oct 16, 2022 at 11:34

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