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Let $\Omega$ be an open bounded domain in $\mathbb{R}^{n}$. We denote $d\left(x\right)$ the distance from $x$ to the boundary of $\Omega$, that is $$ d\left(x\right):=\inf\left\{ \left\Vert x-y\right\Vert :y\in\partial\Omega\right\} . $$ In the book of A.Kufner: Weighted Sobolev spaces, page 50, he claimed that (without any explanation): there always exists $\varepsilon_{0}>0$ such that $$ \int_{\Omega}d^{-\varepsilon_{0}}\left(x\right)dx<\infty. $$

I suspect this is not true. I think at least we need some assumption on the regularity of $\partial\Omega$, for example $\partial\Omega$ is Lipschitz.

My question: is the above statement is true for any open bounded domain?

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No, $d^{-\epsilon}$ can fail to be integrable for all $\epsilon > 0$.


Take $\Omega \subset \mathbb{R}^2$ defined by $$ \Omega = ((0, 1) \times (-1, 1)) \setminus \bigcup_{k = 2}^\infty \{\tfrac{1}{\log(k)}\} \times [0, 1) . $$ If $1/\log(k + 1) < x < 1/\log(k)$ and $0 < y < 1$, then $(x, y) \in \Omega$ and $d(x) < 1/\log(k) - 1/\log(k + 1)$. Thus, $$ \int_\Omega (d(w))^{-\epsilon} dw > \sum_{k = 2}^\infty \left(\frac{1}{\log(k)} - \frac{1}{\log(k + 1)}\right)^{1-\epsilon} , $$ and the series diverges by the ratio test: $$ \left(\frac{1}{\log(k)} - \frac{1}{\log(k + 1)}\right)^{1-\epsilon} \sim \frac{1}{(k \log^2(k))^{1 - \epsilon}} \, . $$

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  • $\begingroup$ The segment $\{0\}\times[0,1)$ should also be taken out to make $\Omega$ an open set. $\endgroup$
    – user111
    Commented Mar 15, 2018 at 19:59
  • $\begingroup$ @user111: If I am not mistaken, it is already not there? $\endgroup$ Commented Mar 15, 2018 at 21:01
  • $\begingroup$ Oh yes, that's right ! $\endgroup$
    – user111
    Commented Mar 15, 2018 at 21:38

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