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Let $E$ be a Banach lattice. Then $u \in E_+$ is said to be a quasi-interior point of $E$ is $$E_u:=\{f \in E:\exists\, c\geq 0 \text{ such that } |f| \leq cu\}$$ is dense in $E.$

Let $\Omega$ be a bounded domain with $C^{\infty}$-boundary and define $u(x)=\operatorname{dist}(x,\partial \Omega).$ Describe the sets $\left(L^p(\Omega)\right)_{u^2} \, (1\leq p <\infty)$ and $\left(C_0(\Omega)\right)_{u}.$

My attempt:

Let $E=L^p(\Omega)$ and $f \in E_{u^2}.$ Then there exists $c\geq 0$ such that $|f| \leq c u^2.$ Therefore $|f|(x) \leq c \operatorname{dist}(x,\partial \Omega)^2$ for all $x.$ Therefore $f \in L^{\infty}(\Omega).$ Therefore $E_{u^2} \subseteq L^{\infty}(\Omega).$ The converse inclusion is obviously true an so $E_{u^2} = L^{\infty}(\Omega).$

For similar reasons, $\left(C_0(\Omega)\right)_{u}=L^{\infty}(\Omega).$

Am I right?

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    $\begingroup$ "The converse inclusion is obviously true" is false. Any ideal of $C_0$ has to consist of continuous functions $\endgroup$
    – erz
    Commented Jun 10, 2020 at 1:29
  • $\begingroup$ "The converse inclusion is obviously true" is also false for $L^p$. For instance, the constant function with value $1$ is not contained in $(L^p(\Omega))_{u^2}$ $\endgroup$
    – Jochen Glueck
    Commented Jun 10, 2020 at 5:30
  • $\begingroup$ Hmm, I'm not quite sure what kind of description you are looking for. We have $(L^p(\Omega))_{u^2} = \{f \in L^p(\Omega): \; \exists c \ge 0: \, |f| \le cu^2\}$ and $(C_0(\Omega))_u = \{f \in C_0(\Omega): \; \exists c \ge 0: \, |f| \le cu\}$ by the very definition of principal ideals, and I find this already quite explicit ;actually, I don't think we'll get anything more explicit than that if we want to describe these principal ideals as subspaces of $L^p(\Omega)$ and $C_0(\Omega)$. [to be continued] $\endgroup$
    – Jochen Glueck
    Commented Jun 10, 2020 at 5:48
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    $\begingroup$ [contiuation] However, we can isometrically isomorphically describe these principal ideals of we endow them with the gauge norm: the space $(L^p(\Omega))_{u^2}$ is isometrically lattice isomorphic (but not equal) to $L^\infty(\Omega)$, and the space $(C_0(\Omega))_u$ is isometrically lattice isomorphic (but of course not equal) to $C_b(\Omega)$. $\endgroup$
    – Jochen Glueck
    Commented Jun 10, 2020 at 5:53
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    $\begingroup$ [continuation] But smoothness up to the boundary can help you to prove that $|f| \le cu^2$ for some number $c \ge 0$: If $f \in C^2(\overline{\Omega})$ and $f$ vanishes on the boundary of $\Omega$ and the normal derivative of $f$ vanishes on the boundary, too, then $|f|$ can indeed by dominated by a multiple of $u^2$ (this is mainly the definition of the derivative - I think the best way to get an intuition for this is to first consider the one-dimensional case $\Omega = (0,1)$). $\endgroup$
    – Jochen Glueck
    Commented Jun 10, 2020 at 9:21

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